\displaystyle \textbf{Question 1: }\text{The probability that it will rain tomorrow is }0.85.\text{ What is the probability}
\displaystyle \text{that it will not rain tomorrow?}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ be the event that it will rain tomorrow.}
\displaystyle \therefore P(A)=0.85
\displaystyle P(\overline{A})=1-P(A)
\displaystyle =1-0.85=0.15
\displaystyle \therefore \text{The probability that it will not rain tomorrow is }0.15.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{A die is thrown. Find the probability of getting:}
\displaystyle \text{(i) a prime number}\hfill\text{[CBSE 2019]}
\displaystyle \text{(ii) }2\text{ or }4
\displaystyle \text{(iii) a multiple of }2\text{ or }3\qquad\text{(iv) an even prime number}\hfill\text{[CBSE 2008]}
\displaystyle \text{(v) a number greater than }5\hfill\text{[CBSE 2008]}
\displaystyle \text{(vi) a number lying between }2\text{ and }6
\displaystyle \text{(vii) a composite number}\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{The possible outcomes on throwing a die are }1,2,3,4,5,6.
\displaystyle \therefore \text{Total number of elementary events}=6.
\displaystyle \text{(i) Prime numbers are }2,3,5.
\displaystyle \therefore \text{Favourable number of elementary events}=3.
\displaystyle \therefore P(\text{getting a prime number})=\frac{3}{6}=\frac{1}{2}.

\displaystyle \text{(ii) Favourable outcomes are }2,4.
\displaystyle \therefore \text{Favourable number of elementary events}=2.
\displaystyle \therefore P(\text{getting }2\text{ or }4)=\frac{2}{6}=\frac{1}{3}.

\displaystyle \text{(iii) Multiples of }2\text{ or }3\text{ are }2,3,4,6.
\displaystyle \therefore \text{Favourable number of elementary events}=4.
\displaystyle \therefore P(\text{getting a multiple of }2\text{ or }3)=\frac{4}{6}=\frac{2}{3}.

\displaystyle \text{(iv) The only even prime number is }2.
\displaystyle \therefore \text{Favourable number of elementary events}=1.
\displaystyle \therefore P(\text{getting an even prime number})=\frac{1}{6}.

\displaystyle \text{(v) The only number greater than }5\text{ is }6.
\displaystyle \therefore \text{Favourable number of elementary events}=1.
\displaystyle \therefore P(\text{getting a number greater than }5)=\frac{1}{6}.

\displaystyle \text{(vi) Numbers lying between }2\text{ and }6\text{ are }3,4,5.
\displaystyle \therefore \text{Favourable number of elementary events}=3.
\displaystyle \therefore P(\text{getting a number lying between }2\text{ and }6)=\frac{3}{6}=\frac{1}{2}.

\displaystyle \text{(vii) Composite numbers are }4,6.
\displaystyle \therefore \text{Favourable number of elementary events}=2.
\displaystyle \therefore P(\text{getting a composite number})=\frac{2}{6}=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Two unbiased dice are thrown. Find the probability that the total of the}
\displaystyle \text{numbers on the dice is greater than }10.\hfill\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{When two unbiased dice are thrown, total number of elementary events}=6\times6=36.
\displaystyle \text{For the total to be greater than }10,\text{ the sum must be }11\text{ or }12.
\displaystyle \text{The favourable outcomes are }(5,6),(6,5),(6,6).
\displaystyle \therefore \text{Favourable number of elementary events}=3.
\displaystyle \therefore P(\text{total greater than }10)=\frac{3}{36}=\frac{1}{12}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A box contains }20\text{ cards numbered from }1\text{ to }20.\text{ A card is drawn at random}
\displaystyle \text{from the box. Find the probability that the number on the drawn card is:}
\displaystyle \text{(i) divisible by }2\text{ or }3\qquad\text{(ii) a prime number}\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=20.
\displaystyle \text{(i) Numbers divisible by }2\text{ or }3\text{ are }2,3,4,6,8,9,10,12,14,15,16,18,20.
\displaystyle \therefore \text{Favourable number of elementary events}=13.
\displaystyle \therefore P(\text{number divisible by }2\text{ or }3)=\frac{13}{20}.

\displaystyle \text{(ii) Prime numbers from }1\text{ to }20\text{ are }2,3,5,7,11,13,17,19.
\displaystyle \therefore \text{Favourable number of elementary events}=8.
\displaystyle \therefore P(\text{getting a prime number})=\frac{8}{20}=\frac{2}{5}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A bag contains }3\text{ red balls, }5\text{ black balls and }4\text{ white balls. A ball is drawn at}
\displaystyle \text{random from the bag. What is the probability that the ball drawn is:}\hfill\text{[CBSE 2008]}
\displaystyle \text{(i) white?}\qquad\text{(ii) red?}\qquad\text{(iii) black?}\qquad\text{(iv) not red?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=3+5+4=12.
\displaystyle \therefore \text{Total number of elementary events}=12.
\displaystyle \text{(i) Number of white balls}=4.
\displaystyle \therefore P(\text{white})=\frac{4}{12}=\frac{1}{3}.

\displaystyle \text{(ii) Number of red balls}=3.
\displaystyle \therefore P(\text{red})=\frac{3}{12}=\frac{1}{4}.

\displaystyle \text{(iii) Number of black balls}=5.
\displaystyle \therefore P(\text{black})=\frac{5}{12}.

\displaystyle \text{(iv) Number of balls which are not red}=5+4=9.
\displaystyle \therefore P(\text{not red})=\frac{9}{12}=\frac{3}{4}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{A bag contains cards numbered from }1\text{ to }49.\text{ A card is drawn from the bag at}
\displaystyle \text{random, after mixing the cards thoroughly. Find the probability that the number on the}
\displaystyle \text{drawn card is:}\hfill\text{[CBSE 2014]}
\displaystyle \text{(i) an odd number}\qquad\text{(ii) a multiple of }5
\displaystyle \text{(iii) a perfect square}\qquad\text{(iv) an even prime number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=49.
\displaystyle \text{(i) There are }25\text{ odd numbers from }1\text{ to }49.
\displaystyle \therefore P(\text{an odd number})=\frac{25}{49}.

\displaystyle \text{(ii) Multiples of }5\text{ are }5,10,15,20,25,30,35,40,45.
\displaystyle \therefore \text{Favourable number of elementary events}=9.
\displaystyle \therefore P(\text{a multiple of }5)=\frac{9}{49}.

\displaystyle \text{(iii) Perfect squares are }1,4,9,16,25,36,49.
\displaystyle \therefore \text{Favourable number of elementary events}=7.
\displaystyle \therefore P(\text{a perfect square})=\frac{7}{49}=\frac{1}{7}.

\displaystyle \text{(iv) The only even prime number is }2.
\displaystyle \therefore \text{Favourable number of elementary events}=1.
\displaystyle \therefore P(\text{an even prime number})=\frac{1}{49}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A piggy bank contains hundred }50\text{ paise coins, fifty Rs. }1\text{ coins, twenty Rs. }2
\displaystyle \text{coins and ten Rs. }5\text{ coins. If it is equally likely that one of the coins will fall out when}
\displaystyle \text{the bank is turned upside down, find the probability that the coin which fell:}
\displaystyle \text{(i) will be a }50\text{ paise coin}\qquad\text{(ii) will be of value more than Rs. }1
\displaystyle \text{(iii) will be of value less than Rs. }5\qquad\text{(iv) will be a Rs. }1\text{ or Rs. }2\text{ coin}
\displaystyle \hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of coins}=100+50+20+10=180.
\displaystyle \therefore \text{Total number of elementary events}=180.
\displaystyle \text{(i) Number of }50\text{ paise coins}=100.
\displaystyle \therefore P(\text{a }50\text{ paise coin})=\frac{100}{180}=\frac{5}{9}.

\displaystyle \text{(ii) Number of coins of value more than Rs. }1=20+10=30.
\displaystyle \therefore P(\text{value more than Rs. }1)=\frac{30}{180}=\frac{1}{6}.

\displaystyle \text{(iii) Number of coins of value less than Rs. }5=100+50+20=170.
\displaystyle \therefore P(\text{value less than Rs. }5)=\frac{170}{180}=\frac{17}{18}.

\displaystyle \text{(iv) Number of Rs. }1\text{ or Rs. }2\text{ coins}=50+20=70.
\displaystyle \therefore P(\text{a Rs. }1\text{ or Rs. }2\text{ coin})=\frac{70}{180}=\frac{7}{18}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Cards numbered }1\text{ to }30\text{ are put in a bag. A card is drawn at random from}
\displaystyle \text{this bag. Find the probability that the number on the drawn card is:}
\displaystyle \text{(i) not divisible by }3\hfill\text{[CBSE 2005]}
\displaystyle \text{(ii) a prime number greater than }7
\displaystyle \text{(iii) not a perfect square number.}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=30.
\displaystyle \text{(i) Numbers divisible by }3\text{ are }3,6,9,12,15,18,21,24,27,30.
\displaystyle \therefore \text{Number of numbers not divisible by }3=30-10=20.
\displaystyle \therefore P(\text{not divisible by }3)=\frac{20}{30}=\frac{2}{3}.

\displaystyle \text{(ii) Prime numbers greater than }7\text{ are }11,13,17,19,23,29.
\displaystyle \therefore \text{Favourable number of elementary events}=6.
\displaystyle \therefore P(\text{a prime number greater than }7)=\frac{6}{30}=\frac{1}{5}.

\displaystyle \text{(iii) Perfect square numbers are }1,4,9,16,25.
\displaystyle \therefore \text{Number of numbers which are not perfect squares}=30-5=25.
\displaystyle \therefore P(\text{not a perfect square number})=\frac{25}{30}=\frac{5}{6}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{A bag contains }4\text{ red, }5\text{ black and }6\text{ white balls. A ball is drawn from the bag}
\displaystyle \text{at random. Find the probability that the ball drawn is:}
\displaystyle \text{(i) white}\qquad\text{(ii) red}\qquad\text{(iii) not black}\qquad\text{(iv) red or white}
\displaystyle \hfill\text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=4+5+6=15.
\displaystyle \therefore \text{Total number of elementary events}=15.
\displaystyle \text{(i) Number of white balls}=6.
\displaystyle \therefore P(\text{white})=\frac{6}{15}=\frac{2}{5}.

\displaystyle \text{(ii) Number of red balls}=4.
\displaystyle \therefore P(\text{red})=\frac{4}{15}.

\displaystyle \text{(iii) Number of balls which are not black}=4+6=10.
\displaystyle \therefore P(\text{not black})=\frac{10}{15}=\frac{2}{3}.

\displaystyle \text{(iv) Number of red or white balls}=4+6=10.
\displaystyle \therefore P(\text{red or white})=\frac{10}{15}=\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{One card is drawn from a well shuffled deck of }52\text{ cards. Find the}
\displaystyle \text{probability of getting:}
\displaystyle \text{(i) a king of red suit}\qquad\text{(ii) a face card}\qquad\text{(iii) a red face card}
\displaystyle \text{(iv) a queen of black suit}\qquad\text{(v) a jack of hearts}\qquad\text{(vi) a spade}
\displaystyle \text{(vii) a queen of hearts}\hfill\text{[CBSE 2024]}\qquad\text{(viii) not a jack}\hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=52.
\displaystyle \text{(i) There are }2\text{ kings of red suit, one each of hearts and diamonds.}
\displaystyle \therefore P(\text{a king of red suit})=\frac{2}{52}=\frac{1}{26}.

\displaystyle \text{(ii) There are }12\text{ face cards, namely }4\text{ kings, }4\text{ queens and }4\text{ jacks.}
\displaystyle \therefore P(\text{a face card})=\frac{12}{52}=\frac{3}{13}.

\displaystyle \text{(iii) There are }6\text{ red face cards, }3\text{ each of hearts and diamonds.}
\displaystyle \therefore P(\text{a red face card})=\frac{6}{52}=\frac{3}{26}.

\displaystyle \text{(iv) There are }2\text{ queens of black suit, one each of spades and clubs.}
\displaystyle \therefore P(\text{a queen of black suit})=\frac{2}{52}=\frac{1}{26}.

\displaystyle \text{(v) There is only one jack of hearts in a deck of }52\text{ cards.}
\displaystyle \therefore P(\text{a jack of hearts})=\frac{1}{52}.

\displaystyle \text{(vi) There are }13\text{ cards of spades in a deck of }52\text{ cards.}
\displaystyle \therefore P(\text{a spade})=\frac{13}{52}=\frac{1}{4}.

\displaystyle \text{(vii) There is only one queen of hearts in a deck of }52\text{ cards.}
\displaystyle \therefore P(\text{a queen of hearts})=\frac{1}{52}.

\displaystyle \text{(viii) There are }4\text{ jacks in a deck of }52\text{ cards.}
\displaystyle \therefore \text{Number of cards which are not jacks}=52-4=48.
\displaystyle \therefore P(\text{not a jack})=\frac{48}{52}=\frac{12}{13}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Five cards - ten, jack, queen, king, and an ace of diamonds are shuffled}
\displaystyle \text{face downwards. One card is picked at random.}
\displaystyle \text{(i) What is the probability that the card is a queen?}
\displaystyle \text{(ii) If a king is drawn first and put aside, what is the probability that the second card picked}
\displaystyle \text{up is (i) ace? \qquad (ii) king?}\hfill\text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{There are }5\text{ cards in all.}
\displaystyle \text{(i) There is only one queen among the }5\text{ cards.}
\displaystyle \therefore P(\text{a queen})=\frac{1}{5}.

\displaystyle \text{(ii) After the king is drawn and put aside, }4\text{ cards remain.}
\displaystyle \text{(i) There is one ace among the remaining }4\text{ cards.}
\displaystyle \therefore P(\text{an ace})=\frac{1}{4}.
\displaystyle \text{(ii) Since the only king has already been put aside, no king remains.}
\displaystyle \therefore P(\text{a king})=\frac{0}{4}=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{A group consists of }12\text{ persons, of which }3\text{ are extremely patient, other }6\text{ are}
\displaystyle \text{extremely honest and the rest are extremely kind. A person from the group is selected at}
\displaystyle \text{random. Assuming that each person is equally likely to be selected, find the probability of}
\displaystyle \text{selecting a person who is (i) extremely patient (ii) extremely kind or honest. Which of the}
\displaystyle \text{above do you prefer more?}\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of persons}=12.
\displaystyle \text{Number of extremely kind persons}=12-3-6=3.
\displaystyle \text{(i) Number of extremely patient persons}=3.
\displaystyle \therefore P(\text{extremely patient})=\frac{3}{12}=\frac{1}{4}.

\displaystyle \text{(ii) Number of extremely kind or honest persons}=3+6=9.
\displaystyle \therefore P(\text{extremely kind or honest})=\frac{9}{12}=\frac{3}{4}.
\displaystyle \text{Since }\frac{3}{4}>\frac{1}{4},\text{ I would prefer a person who is extremely kind or honest.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{A game of chance consists of spinning an arrow which is equally}
\displaystyle \text{likely to come to rest pointing to one of the numbers }1,2,3,\ldots,12
\displaystyle \text{ as shown in the figure. What is the}  \ \text{probability that it will point to:}
\displaystyle \text{(i) }10\text{?}\qquad\text{(ii) an odd number?}\qquad\text{(iii) a number which is multiple of }3\text{?}
\displaystyle \text{(iv) an even number?} \displaystyle \text{Answer:}
\displaystyle \text{The arrow can point to any one of the numbers }1,2,3,\ldots,12.
\displaystyle \therefore \text{Total number of elementary events}=12.
\displaystyle \text{(i) There is only one favourable outcome, namely }10.
\displaystyle \therefore P(\text{pointing to }10)=\frac{1}{12}.

\displaystyle \text{(ii) The odd numbers are }1,3,5,7,9,11.
\displaystyle \therefore \text{Favourable number of elementary events}=6.
\displaystyle \therefore P(\text{an odd number})=\frac{6}{12}=\frac{1}{2}.

\displaystyle \text{(iii) The multiples of }3\text{ are }3,6,9,12.
\displaystyle \therefore \text{Favourable number of elementary events}=4.
\displaystyle \therefore P(\text{a multiple of }3)=\frac{4}{12}=\frac{1}{3}.

\displaystyle \text{(iv) The even numbers are }2,4,6,8,10,12.
\displaystyle \therefore \text{Favourable number of elementary events}=6.
\displaystyle \therefore P(\text{an even number})=\frac{6}{12}=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A box contains cards numbered }3,5,7,9,\ldots,35,37.\text{ A card is drawn at}
\displaystyle \text{random from the box. Find the probability that the number on the drawn card is a prime}
\displaystyle \text{number.}\hfill\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{The cards are numbered with odd numbers from }3\text{ to }37.
\displaystyle \therefore \text{Total number of elementary events}=\frac{37-3}{2}+1=18.
\displaystyle \text{Prime numbers are }3,5,7,11,13,17,19,23,29,31,37.
\displaystyle \therefore \text{Favourable number of elementary events}=11.
\displaystyle \therefore P(\text{a prime number})=\frac{11}{18}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A bag contains }5\text{ red, }8\text{ white and }7\text{ black balls. A ball is drawn at random}
\displaystyle \text{from the bag. Find the probability that the drawn ball is (i) red or white (ii) not black}
\displaystyle \text{(iii) neither white nor black.}\hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls}=5+8+7=20.
\displaystyle \therefore \text{Total number of elementary events}=20.
\displaystyle \text{(i) Number of red or white balls}=5+8=13.
\displaystyle \therefore P(\text{red or white})=\frac{13}{20}.

\displaystyle \text{(ii) Number of balls which are not black}=5+8=13.
\displaystyle \therefore P(\text{not black})=\frac{13}{20}.

\displaystyle \text{(iii) A ball which is neither white nor black must be red.}
\displaystyle \therefore \text{Favourable number of elementary events}=5.
\displaystyle \therefore P(\text{neither white nor black})=\frac{5}{20}=\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the probability that a number selected from the numbers }1\text{ to }25\text{ is not}
\displaystyle \text{a prime number when each of the given numbers is equally likely to be selected.}
\displaystyle \hfill\text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=25.
\displaystyle \text{Prime numbers from }1\text{ to }25\text{ are }2,3,5,7,11,13,17,19,23.
\displaystyle \therefore \text{Number of prime numbers}=9.
\displaystyle \therefore \text{Number of numbers which are not prime}=25-9=16.
\displaystyle \therefore P(\text{not a prime number})=\frac{16}{25}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{A box contains }100\text{ red cards, }200\text{ yellow cards and }50\text{ blue cards. If a}
\displaystyle \text{card is drawn at random from the box, then find the probability that it will be (i) a blue}
\displaystyle \text{card (ii) not a yellow card (iii) neither yellow nor a blue card.}\hfill\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=100+200+50=350.
\displaystyle \therefore \text{Total number of elementary events}=350.
\displaystyle \text{(i) Number of blue cards}=50.
\displaystyle \therefore P(\text{a blue card})=\frac{50}{350}=\frac{1}{7}.

\displaystyle \text{(ii) Number of cards which are not yellow}=100+50=150.
\displaystyle \therefore P(\text{not a yellow card})=\frac{150}{350}=\frac{3}{7}.

\displaystyle \text{(iii) A card which is neither yellow nor blue must be red.}
\displaystyle \therefore \text{Favourable number of elementary events}=100.
\displaystyle \therefore P(\text{neither yellow nor blue})=\frac{100}{350}=\frac{2}{7}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the probability that a number selected at random from the numbers }
\displaystyle 1,2,3,\ldots,35 \ \text{is a (i) prime number (ii) multiple of }7\text{ (iii) a multiple of }3\text{ or }5.
\displaystyle \hfill\text{[CBSE 2006C]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=35.
\displaystyle \text{(i) Prime numbers are }2,3,5,7,11,13,17,19,23,29,31.
\displaystyle \therefore \text{Favourable number of elementary events}=11.
\displaystyle \therefore P(\text{a prime number})=\frac{11}{35}.

\displaystyle \text{(ii) Multiples of }7\text{ are }7,14,21,28,35.
\displaystyle \therefore \text{Favourable number of elementary events}=5.
\displaystyle \therefore P(\text{a multiple of }7)=\frac{5}{35}=\frac{1}{7}.

\displaystyle \text{(iii) Multiples of }3\text{ are }3,6,9,12,15,18,21,24,27,30,33.
\displaystyle \text{Multiples of }5\text{ are }5,10,15,20,25,30,35.
\displaystyle \text{The common multiples of }3\text{ and }5\text{ are }15,30.
\displaystyle \therefore \text{Favourable number of elementary events}=11+7-2=16.
\displaystyle \therefore P(\text{a multiple of }3\text{ or }5)=\frac{16}{35}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A bag contains tickets numbered }11,12,13,\ldots,30.\text{ A ticket is taken out from}
\displaystyle \text{the bag at random. Find the probability that the number on the drawn ticket (i) is a}
\displaystyle \text{multiple of }7\text{ (ii) is greater than }15\text{ and a multiple of }5.\hfill\text{[CBSE 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=30-11+1=20.
\displaystyle \text{(i) Multiples of }7\text{ from }11\text{ to }30\text{ are }14,21,28.
\displaystyle \therefore \text{Favourable number of elementary events}=3.
\displaystyle \therefore P(\text{a multiple of }7)=\frac{3}{20}.

\displaystyle \text{(ii) Numbers greater than }15\text{ and multiples of }5\text{ are }20,25,30.
\displaystyle \therefore \text{Favourable number of elementary events}=3.
\displaystyle \therefore P(\text{greater than }15\text{ and a multiple of }5)=\frac{3}{20}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{A bag contains lemon flavoured candies only. Malini takes out one }
\displaystyle \text{candy without looking into the bag. What is the probability that she takes out:}
\displaystyle \text{(i) an orange flavoured candy?}\qquad\text{(ii) a lemon flavoured candy?}
\displaystyle \text{Answer:}
\displaystyle \text{Since the bag contains only lemon flavoured candies,}
\displaystyle \text{(i) Getting an orange flavoured candy is an impossible event.}
\displaystyle \therefore P(\text{an orange flavoured candy})=0.

\displaystyle \text{(ii) Getting a lemon flavoured candy is a sure event.}
\displaystyle \therefore P(\text{a lemon flavoured candy})=1.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{It is given that in a group of }3\text{ students, the probability of }2\text{ students not}
\displaystyle \text{having the same birthday is }0.992.\text{ What is the probability that the }2\text{ students have}
\displaystyle \text{the same birthday?}
\displaystyle \text{Answer:}
\displaystyle P(\text{not having the same birthday})=0.992.
\displaystyle P(\text{having the same birthday})=1-P(\text{not having the same birthday}).
\displaystyle =1-0.992=0.008.
\displaystyle \therefore \text{The probability that the }2\text{ students have the same birthday is }0.008.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Cards marked with numbers }13,14,15,\ldots,60\text{ are placed in a box and mixed}
\displaystyle \text{thoroughly. One card is drawn at random from the box. Find the probability that the number}
\displaystyle \text{on the card drawn is (i) divisible by }5\text{ (ii) a perfect square.}\hfill\text{[CBSE 2007]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=60-13+1=48.
\displaystyle \text{(i) Numbers divisible by }5\text{ are }15,20,25,30,35,40,45,50,55,60.
\displaystyle \therefore \text{Favourable number of elementary events}=10.
\displaystyle \therefore P(\text{a number divisible by }5)=\frac{10}{48}=\frac{5}{24}.

\displaystyle \text{(ii) Perfect square numbers from }13\text{ to }60\text{ are }16,25,36,49.
\displaystyle \therefore \text{Favourable number of elementary events}=4.
\displaystyle \therefore P(\text{a perfect square})=\frac{4}{48}=\frac{1}{12}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Harpreet tosses two different coins simultaneously (say, one is of Rs. }1\text{ and}
\displaystyle \text{the other of Rs. }2\text{). What is the probability that he gets at least one head?}
\displaystyle \text{Answer:}
\displaystyle \text{When two different coins are tossed, the possible outcomes are }HH,HT,TH,TT.
\displaystyle \therefore \text{Total number of elementary events}=4.
\displaystyle \text{The outcomes favourable to getting at least one head are }HH,HT,TH.
\displaystyle \therefore \text{Favourable number of elementary events}=3.
\displaystyle \therefore P(\text{at least one head})=\frac{3}{4}.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A lot consists of }144\text{ ball pens of which }20\text{ are defective and others are good.}
\displaystyle \text{Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws}
\displaystyle \text{one pen at random and gives it to her. What is the probability that:}
\displaystyle \text{(i) She will buy it?}\qquad\text{(ii) She will not buy it?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of pens}=144.
\displaystyle \text{Number of defective pens}=20.
\displaystyle \therefore \text{Number of good pens}=144-20=124.
\displaystyle \text{(i) Nuri will buy the pen if it is good.}
\displaystyle \therefore P(\text{she will buy it})=\frac{124}{144}=\frac{31}{36}.

\displaystyle \text{(ii) Nuri will not buy the pen if it is defective.}
\displaystyle \therefore P(\text{she will not buy it})=\frac{20}{144}=\frac{5}{36}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{12 defective pens are accidentally mixed with }132\text{ good ones. It is not possible}
\displaystyle \text{to just look at a pen and tell whether or not it is defective. One pen is taken out at random}
\displaystyle \text{from this lot. Determine the probability that the pen taken out is a good one.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of pens}=12+132=144.
\displaystyle \text{Number of good pens}=132.
\displaystyle \therefore P(\text{a good pen})=\frac{132}{144}=\frac{11}{12}.
\displaystyle \therefore \text{The probability that the pen taken out is a good one is }\frac{11}{12}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Examine each of the following statements and comment:}
\displaystyle \text{(i) If two coins are tossed at the same time, there are }3\text{ possible outcomes -two heads, two}
\displaystyle \text{tails, or one of each. Therefore, for each outcome, the probability of occurrence is }\frac{1}{3}.
\displaystyle \text{(ii) If a die is thrown once, there are two possible outcomes - an odd number or an even}
\displaystyle \text{number. Therefore, the probability of obtaining an odd number is }\frac{1}{2}\text{ and the probability}
\displaystyle \text{of obtaining an even number is }\frac{1}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) The statement is incorrect.}
\displaystyle \text{When two coins are tossed, the equally likely elementary outcomes are }HH,HT,TH,TT.
\displaystyle \therefore \text{Total number of elementary events}=4.
\displaystyle P(\text{two heads})=P(HH)=\frac{1}{4}.
\displaystyle P(\text{two tails})=P(TT)=\frac{1}{4}.
\displaystyle P(\text{one of each})=P(HT\text{ or }TH)=\frac{2}{4}=\frac{1}{2}.
\displaystyle \therefore \text{The three stated outcomes are not equally likely.}

\displaystyle \text{(ii) The statement is correct.}
\displaystyle \text{The possible outcomes on throwing a die are }1,2,3,4,5,6.
\displaystyle \text{The odd numbers are }1,3,5\text{ and the even numbers are }2,4,6.
\displaystyle \therefore P(\text{an odd number})=\frac{3}{6}=\frac{1}{2}.
\displaystyle \therefore P(\text{an even number})=\frac{3}{6}=\frac{1}{2}.
\displaystyle \therefore \text{The probabilities stated in (ii) are correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In a simultaneous throw of a pair of dice, find the probability of getting:}
\displaystyle \text{(i) }8\text{ as the sum}\qquad\text{(ii) }2\text{ will not come either time}\hfill\text{[CBSE 2015, 19]}
\displaystyle \text{(iii) a doublet of prime numbers}\qquad\text{(iv) a doublet of odd numbers}\hfill\text{[CBSE 2017]}
\displaystyle \text{(v) a sum greater than }9\qquad\text{(vi) an even number on first}
\displaystyle \text{(vii) an even number on one and a multiple of }3\text{ on the other}
\displaystyle \text{(viii) neither }9\text{ nor }11\text{ as the sum of the numbers on the faces}
\displaystyle \text{(ix) a sum less than }6\qquad\text{(x) a sum less than }7
\displaystyle \text{(xi) a sum more than }7\qquad\text{(xii) }5\text{ at least once}\hfill\text{[CBSE 2019]}
\displaystyle \text{(xiii) a number other than }5\text{ on any dice}\qquad\text{(xiv) even number on each die}
\displaystyle \hfill\text{[CBSE 2014, 15]}
\displaystyle \text{(xv) }5\text{ as the sum}\hfill\text{[CBSE 2014, 15]}\qquad\text{(xvi) }2\text{ will come up at least once}
\displaystyle \hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{When two dice are thrown, total number of elementary events}=6\times6=36.
\displaystyle \text{(i) The favourable outcomes are }(2,6),(3,5),(4,4),(5,3),(6,2).
\displaystyle \therefore P(\text{sum }8)=\frac{5}{36}.

\displaystyle \text{(ii) For }2\text{ not to occur on either die, each die can show }1,3,4,5\text{ or }6.
\displaystyle \therefore \text{Favourable number of elementary events}=5\times5=25.
\displaystyle \therefore P(\text{2 does not occur on either die})=\frac{25}{36}.

\displaystyle \text{(iii) The favourable doublets of prime numbers are }(2,2),(3,3),(5,5).
\displaystyle \therefore P(\text{a doublet of prime numbers})=\frac{3}{36}=\frac{1}{12}.

\displaystyle \text{(iv) The favourable doublets of odd numbers are }(1,1),(3,3),(5,5).
\displaystyle \therefore P(\text{a doublet of odd numbers})=\frac{3}{36}=\frac{1}{12}.

\displaystyle \text{(v) A sum greater than }9\text{ means a sum of }10,11\text{ or }12.
\displaystyle \text{The favourable outcomes are }(4,6),(5,5),(6,4),(5,6),(6,5),(6,6).
\displaystyle \therefore P(\text{sum greater than }9)=\frac{6}{36}=\frac{1}{6}.

\displaystyle \text{(vi) The first die can show any one of }2,4,6\text{ and the second any number.}
\displaystyle \therefore \text{Favourable number of elementary events}=3\times6=18.
\displaystyle \therefore P(\text{an even number on first die})=\frac{18}{36}=\frac{1}{2}.

\displaystyle \text{(vii) Even numbers are }2,4,6\text{ and multiples of }3\text{ are }3,6.
\displaystyle \text{Even on first and multiple of }3\text{ on second gives }3\times2=6\text{ outcomes.}
\displaystyle \text{Multiple of }3\text{ on first and even on second also gives }2\times3=6\text{ outcomes.}
\displaystyle \text{The outcome }(6,6)\text{ is common to both cases and is counted only once.}
\displaystyle \therefore \text{Favourable number of elementary events}=6+6-1=11.
\displaystyle \therefore P(\text{even on one and multiple of }3\text{ on the other})=\frac{11}{36}.

\displaystyle \text{(viii) Number of outcomes with sum }9=4\text{ and with sum }11=2.
\displaystyle \therefore \text{Number of outcomes with neither sum }9\text{ nor }11=36-(4+2)=30.
\displaystyle \therefore P(\text{neither }9\text{ nor }11\text{ as the sum})=\frac{30}{36}=\frac{5}{6}.

\displaystyle \text{(ix) For a sum less than }6,\text{ possible sums are }2,3,4,5.
\displaystyle \therefore \text{Favourable number of elementary events}=1+2+3+4=10.
\displaystyle \therefore P(\text{sum less than }6)=\frac{10}{36}=\frac{5}{18}.

\displaystyle \text{(x) For a sum less than }7,\text{ possible sums are }2,3,4,5,6.
\displaystyle \therefore \text{Favourable number of elementary events}=1+2+3+4+5=15.
\displaystyle \therefore P(\text{sum less than }7)=\frac{15}{36}=\frac{5}{12}.

\displaystyle \text{(xi) For a sum more than }7,\text{ possible sums are }8,9,10,11,12.
\displaystyle \therefore \text{Favourable number of elementary events}=5+4+3+2+1=15.
\displaystyle \therefore P(\text{sum more than }7)=\frac{15}{36}=\frac{5}{12}.

\displaystyle \text{(xii) If }5\text{ occurs at least once, the favourable outcomes are}
\displaystyle (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5).
\displaystyle \therefore \text{Favourable number of elementary events}=11.
\displaystyle \therefore P(\text{5 at least once})=\frac{11}{36}.

\displaystyle \text{(xiii) For a number other than }5\text{ on either die, each die has }5\text{ choices.}
\displaystyle \therefore \text{Favourable number of elementary events}=5\times5=25.
\displaystyle \therefore P(\text{a number other than }5\text{ on either die})=\frac{25}{36}.

\displaystyle \text{(xiv) Each die can show any one of the even numbers }2,4,6.
\displaystyle \therefore \text{Favourable number of elementary events}=3\times3=9.
\displaystyle \therefore P(\text{an even number on each die})=\frac{9}{36}=\frac{1}{4}.

\displaystyle \text{(xv) The favourable outcomes for sum }5\text{ are }(1,4),(2,3),(3,2),(4,1).
\displaystyle \therefore P(\text{sum }5)=\frac{4}{36}=\frac{1}{9}.

\displaystyle \text{(xvi) If }2\text{ occurs at least once, the favourable outcomes are}
\displaystyle (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(1,2),(3,2),(4,2),(5,2),(6,2).
\displaystyle \therefore \text{Favourable number of elementary events}=11.
\displaystyle \therefore P(\text{2 at least once})=\frac{11}{36}.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{What is the probability that a leap year has }53\text{ Tuesdays and }53
\displaystyle \text{Mondays?}\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days}=52\text{ weeks}+2\text{ days.}
\displaystyle \text{Thus, every weekday occurs }52\text{ times and two consecutive weekdays occur }53\text{ times.}
\displaystyle \text{The possible pairs of extra days are:}
\displaystyle (\text{Sun, Mon}),(\text{Mon, Tue}),(\text{Tue, Wed}),(\text{Wed, Thu}),
\displaystyle (\text{Thu, Fri}),(\text{Fri, Sat}),(\text{Sat, Sun}).
\displaystyle \therefore \text{Total number of elementary events}=7.
\displaystyle \text{For }53\text{ Mondays and }53\text{ Tuesdays, the extra days must be Monday and Tuesday.}
\displaystyle \therefore \text{Favourable number of elementary events}=1.
\displaystyle \therefore P(53\text{ Mondays and }53\text{ Tuesdays})=\frac{1}{7}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{A black die and a white die are thrown at the same time. Write all the}
\displaystyle \text{possible outcomes. What is the probability that:}
\displaystyle \text{(i) the sum of the two numbers that turn up is }8\text{?}
\displaystyle \text{(ii) of obtaining a total of }6\text{?}\hfill\text{[CBSE 2014]}
\displaystyle \text{(iii) of obtaining a total of }10\text{?}\hfill\text{[CBSE 2014]}
\displaystyle \text{(iv) of obtaining the same number on both dice?}
\displaystyle \text{(v) of obtaining a total more than }9\text{?}
\displaystyle \text{(vi) the sum of the two numbers appearing on the top of the dice is }13\text{?}
\displaystyle \text{(vii) the sum of the numbers appearing on the top of the dice is less than or equal to }12\text{?}
\displaystyle \text{(viii) the product of numbers appearing on the top of the dice is less than }9\text{?}\hfill\text{[CBSE 2014]}
\displaystyle \text{(ix) the difference of the numbers appearing on the top of the two dice is }2\text{?}\hfill\text{[CBSE 2014]}
\displaystyle \text{(x) the numbers obtained have a product less than }16\text{?}\hfill\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first number denote the black die and the second number the white die.}
\displaystyle \text{The possible outcomes are:}
\displaystyle (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
\displaystyle (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
\displaystyle (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),
\displaystyle (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),
\displaystyle (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),
\displaystyle (6,1),(6,2),(6,3),(6,4),(6,5),(6,6).
\displaystyle \therefore \text{Total number of elementary events}=36.
\displaystyle \text{(i) Favourable outcomes are }(2,6),(3,5),(4,4),(5,3),(6,2).
\displaystyle \therefore P(\text{sum }8)=\frac{5}{36}.

\displaystyle \text{(ii) Favourable outcomes are }(1,5),(2,4),(3,3),(4,2),(5,1).
\displaystyle \therefore P(\text{total }6)=\frac{5}{36}.

\displaystyle \text{(iii) Favourable outcomes are }(4,6),(5,5),(6,4).
\displaystyle \therefore P(\text{total }10)=\frac{3}{36}=\frac{1}{12}.

\displaystyle \text{(iv) Favourable outcomes are }(1,1),(2,2),(3,3),(4,4),(5,5),(6,6).
\displaystyle \therefore P(\text{same number on both dice})=\frac{6}{36}=\frac{1}{6}.

\displaystyle \text{(v) A total more than }9\text{ means a total of }10,11\text{ or }12.
\displaystyle \text{Favourable outcomes are }(4,6),(5,5),(6,4),(5,6),(6,5),(6,6).
\displaystyle \therefore P(\text{total more than }9)=\frac{6}{36}=\frac{1}{6}.

\displaystyle \text{(vi) The maximum possible sum of the numbers on two dice is }6+6=12.
\displaystyle \therefore \text{Getting a sum of }13\text{ is an impossible event.}
\displaystyle \therefore P(\text{sum }13)=0.

\displaystyle \text{(vii) The sum of the numbers on two dice is always less than or equal to }12.
\displaystyle \therefore \text{It is a sure event.}
\displaystyle \therefore P(\text{sum less than or equal to }12)=1.

\displaystyle \text{(viii) The favourable outcomes for a product less than }9\text{ are:}
\displaystyle (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
\displaystyle (2,1),(2,2),(2,3),(2,4),(3,1),(3,2),
\displaystyle (4,1),(4,2),(5,1),(6,1).
\displaystyle \therefore \text{Favourable number of elementary events}=16.
\displaystyle \therefore P(\text{product less than }9)=\frac{16}{36}=\frac{4}{9}.

\displaystyle \text{(ix) Favourable outcomes are }(1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4).
\displaystyle \therefore \text{Favourable number of elementary events}=8.
\displaystyle \therefore P(\text{difference }2)=\frac{8}{36}=\frac{2}{9}.

\displaystyle \text{(x) The favourable outcomes for a product less than }16\text{ are:}
\displaystyle (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
\displaystyle (2,1),(2,2),(2,3),(2,4),(2,5),(2,6),
\displaystyle (3,1),(3,2),(3,3),(3,4),(3,5),
\displaystyle (4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(6,1),(6,2).
\displaystyle \therefore \text{Favourable number of elementary events}=25.
\displaystyle \therefore P(\text{product less than }16)=\frac{25}{36}.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{All red face cards are removed from a pack of playing cards. The remaining}
\displaystyle \text{cards are well shuffled and then a card is drawn at random from them. Find the probability}
\displaystyle \text{that the drawn card is (i) a red card (ii) a face card and (iii) a card of clubs.}
\displaystyle \hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{There are }6\text{ red face cards, namely the jack, queen and king of hearts and diamonds.}
\displaystyle \therefore \text{Number of cards remaining}=52-6=46.
\displaystyle \text{(i) Number of red cards remaining}=26-6=20.
\displaystyle \therefore P(\text{a red card})=\frac{20}{46}=\frac{10}{23}.

\displaystyle \text{(ii) The remaining face cards are the }6\text{ black face cards.}
\displaystyle \therefore P(\text{a face card})=\frac{6}{46}=\frac{3}{23}.

\displaystyle \text{(iii) No card of clubs has been removed, so all }13\text{ club cards remain.}
\displaystyle \therefore P(\text{a card of clubs})=\frac{13}{46}.
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{In a bag there are }44\text{ identical cards with figure of circle or square on them. There}
\displaystyle \text{are }24\text{ circles, of which }9\text{ are blue and the rest are green, and }20\text{ squares of which}
\displaystyle 11\text{ are blue and the rest are green. One card is drawn from the bag at random. Find the}
\displaystyle \text{probability that it has the figure of (i) square (ii) green colour (iii) blue circle and}
\displaystyle \text{(iv) green square.}\hfill\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=44.
\displaystyle \text{Number of green circles}=24-9=15.
\displaystyle \text{Number of green squares}=20-11=9.
\displaystyle \text{(i) Number of square cards}=20.
\displaystyle \therefore P(\text{square})=\frac{20}{44}=\frac{5}{11}.

\displaystyle \text{(ii) Number of green cards}=15+9=24.
\displaystyle \therefore P(\text{green colour})=\frac{24}{44}=\frac{6}{11}.

\displaystyle \text{(iii) Number of blue circle cards}=9.
\displaystyle \therefore P(\text{blue circle})=\frac{9}{44}.

\displaystyle \text{(iv) Number of green square cards}=9.
\displaystyle \therefore P(\text{green square})=\frac{9}{44}.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{All kings and queens are removed from a pack of }52\text{ cards. The remaining cards}
\displaystyle \text{are well-shuffled and then a card is randomly drawn from it. Find the probability that this}
\displaystyle \text{card is (i) a red face card (ii) a black card}\hfill\text{[CBSE 2014]}
\displaystyle \text{(iii) either a red card or a queen}\qquad\text{(iv) red and a queen}
\displaystyle \text{Answer:}
\displaystyle \text{There are }4\text{ kings and }4\text{ queens in a pack of }52\text{ cards.}
\displaystyle \therefore \text{Number of cards removed}=4+4=8.
\displaystyle \therefore \text{Number of cards remaining}=52-8=44.
\displaystyle \text{(i) After removing all kings and queens, only jacks remain as face cards.}
\displaystyle \text{There are }2\text{ red jacks, one each of hearts and diamonds.}
\displaystyle \therefore P(\text{a red face card})=\frac{2}{44}=\frac{1}{22}.

\displaystyle \text{(ii) Of the }8\text{ cards removed, }4\text{ are black.}
\displaystyle \therefore \text{Number of black cards remaining}=26-4=22.
\displaystyle \therefore P(\text{a black card})=\frac{22}{44}=\frac{1}{2}.

\displaystyle \text{(iii) Since all queens have been removed, no queen remains in the pack.}
\displaystyle \text{Number of red cards remaining}=26-4=22.
\displaystyle \therefore P(\text{either a red card or a queen})=\frac{22}{44}=\frac{1}{2}.

\displaystyle \text{(iv) Since all queens have been removed, getting a red queen is impossible.}
\displaystyle \therefore P(\text{red and a queen})=0.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The king, queen and jack of clubs are removed from a deck of }52\text{ playing cards}
\displaystyle \text{and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the}
\displaystyle \text{probability of getting:}
\displaystyle \text{(i) a card of heart}\qquad\text{(ii) a queen}\qquad\text{(iii) a card of clubs}
\displaystyle \text{(iv) a face card}\qquad\text{(v) a queen of diamond.}\hfill\text{[CBSE 2009, 17]}
\displaystyle \text{Answer:}
\displaystyle \text{Number of cards remaining}=52-3=49.
\displaystyle \therefore \text{Total number of elementary events}=49.
\displaystyle \text{(i) No heart card has been removed, so all }13\text{ heart cards remain.}
\displaystyle \therefore P(\text{a card of heart})=\frac{13}{49}.

\displaystyle \text{(ii) One queen has been removed, so }3\text{ queens remain.}
\displaystyle \therefore P(\text{a queen})=\frac{3}{49}.

\displaystyle \text{(iii) Three club cards have been removed from the }13\text{ club cards.}
\displaystyle \therefore \text{Number of club cards remaining}=13-3=10.
\displaystyle \therefore P(\text{a card of clubs})=\frac{10}{49}.

\displaystyle \text{(iv) A deck has }12\text{ face cards. Three face cards have been removed.}
\displaystyle \therefore \text{Number of face cards remaining}=12-3=9.
\displaystyle \therefore P(\text{a face card})=\frac{9}{49}.

\displaystyle \text{(v) The queen of diamonds has not been removed.}
\displaystyle \therefore \text{Favourable number of elementary events}=1.
\displaystyle \therefore P(\text{a queen of diamond})=\frac{1}{49}.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{(i) A lot of }20\text{ bulbs contain }4\text{ defective ones. One bulb is drawn at}
\displaystyle \text{random from the lot. What is the probability that this bulb is defective?}
\displaystyle \text{(ii) Suppose the bulb drawn in (i) is not defective and not replaced. Now a bulb is drawn}
\displaystyle \text{at random from the rest. What is the probability that this bulb is not defective?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Total number of bulbs}=20.
\displaystyle \text{Number of defective bulbs}=4.
\displaystyle \therefore P(\text{a defective bulb})=\frac{4}{20}=\frac{1}{5}.

\displaystyle \text{(ii) Initially, number of non-defective bulbs}=20-4=16.
\displaystyle \text{Since one non-defective bulb is drawn and not replaced,}
\displaystyle \text{number of bulbs remaining}=20-1=19.
\displaystyle \text{Number of non-defective bulbs remaining}=16-1=15.
\displaystyle \therefore P(\text{a non-defective bulb})=\frac{15}{19}.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{A box contains }90\text{ discs which are numbered from }1\text{ to }90.\text{ If one disc is}
\displaystyle \text{drawn at random from the box, find the probability that it bears (i) a two digit number}
\displaystyle \text{(ii) a perfect square number (iii) a number divisible by }5.
\displaystyle \hfill\text{[CBSE 2017, 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=90.
\displaystyle \text{(i) The two digit numbers are from }10\text{ to }90.
\displaystyle \therefore \text{Number of two digit numbers}=90-10+1=81.
\displaystyle \therefore P(\text{a two digit number})=\frac{81}{90}=\frac{9}{10}.

\displaystyle \text{(ii) Perfect square numbers are }1,4,9,16,25,36,49,64,81.
\displaystyle \therefore \text{Favourable number of elementary events}=9.
\displaystyle \therefore P(\text{a perfect square number})=\frac{9}{90}=\frac{1}{10}.

\displaystyle \text{(iii) The numbers divisible by }5\text{ are }5,10,15,\ldots,90.
\displaystyle \therefore \text{Favourable number of elementary events}=\frac{90}{5}=18.
\displaystyle \therefore P(\text{a number divisible by }5)=\frac{18}{90}=\frac{1}{5}.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Two dice, one blue and one grey, are thrown at the same time. Complete the}
\displaystyle \text{following table:}
\displaystyle \begin{array}{c|ccccccccccc}\text{Sum on two dice}&2&3&4&5&6&7&8&9&10&11&12\\ \hline \text{Probability}&&&&&&&&&&&\end{array}
\displaystyle \text{From the above table a student argues that there are }11\text{ possible outcomes }2,3,4,5,6,7,8,9,
\displaystyle 10,11\text{ and }12.\text{ Therefore, each of them has probability }\frac{1}{11}.\text{ Do you agree with this argument?}
\displaystyle \text{Answer:}
\displaystyle \text{When two dice are thrown, total number of elementary events}=6\times6=36.
\displaystyle \text{The numbers of ways of obtaining the sums }2,3,\ldots,12\text{ are respectively}
\displaystyle 1,2,3,4,5,6,5,4,3,2,1.
\displaystyle \therefore \begin{array}{c|ccccccccccc}\text{Sum}&2&3&4&5&6&7&8&9&10&11&12\\ \hline \text{Probability}&\frac{1}{36}&\frac{2}{36}&\frac{3}{36}&\frac{4}{36}&\frac{5}{36}&\frac{6}{36}&\frac{5}{36}&\frac{4}{36}&\frac{3}{36}&\frac{2}{36}&\frac{1}{36}\end{array}
\displaystyle \text{The student's argument is incorrect because the different sums are not equally likely.}
\displaystyle \text{For example, the sum }2\text{ occurs only as }(1,1),\text{ whereas the sum }7\text{ occurs as}
\displaystyle (1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
\displaystyle \therefore P(\text{sum }2)=\frac{1}{36}\text{ and }P(\text{sum }7)=\frac{6}{36}=\frac{1}{6}.
\displaystyle \therefore \text{The probabilities of the sums are not all equal to }\frac{1}{11}.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Three unbiased coins are tossed simultaneously. Find the probability of}
\displaystyle \text{getting: (i) at least one head (ii) exactly one tail (iii) two heads and one tail}
\displaystyle \text{(iv) at most two heads.}\hfill\text{[CBSE 2024, 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{The possible outcomes are }HHH,HHT,HTH,THH,HTT,THT,TTH,TTT.
\displaystyle \therefore \text{Total number of elementary events}=8.
\displaystyle \text{(i) Only }TTT\text{ has no head.}
\displaystyle \therefore P(\text{at least one head})=1-\frac{1}{8}=\frac{7}{8}.

\displaystyle \text{(ii) The favourable outcomes are }HHT,HTH,THH.
\displaystyle \therefore P(\text{exactly one tail})=\frac{3}{8}.

\displaystyle \text{(iii) The favourable outcomes are }HHT,HTH,THH.
\displaystyle \therefore P(\text{two heads and one tail})=\frac{3}{8}.

\displaystyle \text{(iv) At most two heads means }0,1\text{ or }2\text{ heads.}
\displaystyle \text{Only }HHH\text{ has more than two heads.}
\displaystyle \therefore P(\text{at most two heads})=1-\frac{1}{8}=\frac{7}{8}.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{The numbers on a die are replaced by the first six even numbers.}
\displaystyle \text{The die is rolled once. Find the probability that the number appearing on the die is:}
\displaystyle \text{(i) greater than }4\qquad\text{(ii) divisible by }3\qquad\text{(iii) not a multiple of }10
\displaystyle \hfill\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{The numbers on the die are }2,4,6,8,10,12.
\displaystyle \therefore \text{Total number of elementary events}=6.
\displaystyle \text{(i) Numbers greater than }4\text{ are }6,8,10,12.
\displaystyle \therefore P(\text{greater than }4)=\frac{4}{6}=\frac{2}{3}.

\displaystyle \text{(ii) Numbers divisible by }3\text{ are }6,12.
\displaystyle \therefore P(\text{divisible by }3)=\frac{2}{6}=\frac{1}{3}.

\displaystyle \text{(iii) The only multiple of }10\text{ is }10.
\displaystyle \therefore \text{Numbers which are not multiples of }10\text{ are }2,4,6,8,12.
\displaystyle \therefore P(\text{not a multiple of }10)=\frac{5}{6}.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }65\%\text{ of the population has black eyes, }25\%\text{ have brown eyes and the}
\displaystyle \text{remaining have blue eyes, what is the probability that a person selected at random has:}
\displaystyle \text{(i) blue eyes}\qquad\text{(ii) brown or black eyes?}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle P(\text{black eyes})=\frac{65}{100}=0.65.
\displaystyle P(\text{brown eyes})=\frac{25}{100}=0.25.
\displaystyle \text{(i) Percentage of people having blue eyes}=100\%-65\%-25\%=10\%.
\displaystyle \therefore P(\text{blue eyes})=\frac{10}{100}=0.10.

\displaystyle \text{(ii) Percentage of people having brown or black eyes}=25\%+65\%=90\%.
\displaystyle \therefore P(\text{brown or black eyes})=\frac{90}{100}=0.90.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Saima and Arya were born in the month of June in the year }2012.\text{ Find the}
\displaystyle \text{probability that: (i) they have different dates of birth (ii) they have same date of birth.}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{June has }30\text{ days. Therefore, each can have a birthday on any of the }30\text{ days.}
\displaystyle \therefore \text{Total number of elementary events}=30\times30=900.
\displaystyle \text{(i) For different dates, Saima can have any of }30\text{ dates and Arya any of the remaining }29.
\displaystyle \therefore \text{Favourable number of elementary events}=30\times29=870.
\displaystyle \therefore P(\text{different dates of birth})=\frac{870}{900}=\frac{29}{30}.

\displaystyle \text{(ii) They can have the same date of birth in }30\text{ ways.}
\displaystyle \therefore \text{Favourable number of elementary events}=30.
\displaystyle \therefore P(\text{same date of birth})=\frac{30}{900}=\frac{1}{30}.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{All face cards of spades are removed from a pack of }52\text{ playing cards and the}
\displaystyle \text{remaining pack is shuffled. A card is then drawn at random from the remaining pack. Find}
\displaystyle \text{the probability of getting: (i) a face card (ii) an ace or a jack.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{The face cards of spades are the jack, queen and king of spades.}
\displaystyle \therefore \text{Number of cards removed}=3.
\displaystyle \therefore \text{Number of cards remaining}=52-3=49.
\displaystyle \text{(i) Originally, there are }12\text{ face cards. Three face cards have been removed.}
\displaystyle \therefore \text{Number of face cards remaining}=12-3=9.
\displaystyle \therefore P(\text{a face card})=\frac{9}{49}.

\displaystyle \text{(ii) All }4\text{ aces remain, while the jack of spades has been removed.}
\displaystyle \therefore \text{Number of jacks remaining}=4-1=3.
\displaystyle \therefore \text{Number of favourable cards}=4+3=7.
\displaystyle \therefore P(\text{an ace or a jack})=\frac{7}{49}=\frac{1}{7}.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Renu and Simran were born in the year }2000\text{ which is a leap year. Find the}
\displaystyle \text{probability that: (i) both have same birthday (ii) both have different birthdays.}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days.}
\displaystyle \therefore \text{Total number of possible ordered pairs of birthdays}=366\times366.
\displaystyle \text{(i) They can have the same birthday in }366\text{ ways.}
\displaystyle \therefore P(\text{same birthday})=\frac{366}{366\times366}=\frac{1}{366}.

\displaystyle \text{(ii) }P(\text{different birthdays})=1-P(\text{same birthday}).
\displaystyle =1-\frac{1}{366}=\frac{365}{366}.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{While shuffling a pack of }52\text{ cards, one card was accidentally dropped. Find the}
\displaystyle \text{probability that the dropped card (i) is not a face card (ii) is a black king.}
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=52.
\displaystyle \text{(i) There are }12\text{ face cards in a pack of }52\text{ cards.}
\displaystyle \therefore \text{Number of cards which are not face cards}=52-12=40.
\displaystyle \therefore P(\text{not a face card})=\frac{40}{52}=\frac{10}{13}.

\displaystyle \text{(ii) There are }2\text{ black kings, namely the king of spades and king of clubs.}
\displaystyle \therefore P(\text{a black king})=\frac{2}{52}=\frac{1}{26}.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{All the face cards are removed from the pack of }52\text{ cards and a card is}
\displaystyle \text{drawn at random from the remaining cards. Find the probability that the card so drawn is}
\displaystyle \text{(i) a spade \qquad (ii) not an ace.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{There are }12\text{ face cards in a pack of }52\text{ cards.}
\displaystyle \therefore \text{Number of cards remaining}=52-12=40.
\displaystyle \text{(i) Of the }13\text{ spades, the jack, queen and king of spades are removed.}
\displaystyle \therefore \text{Number of spades remaining}=13-3=10.
\displaystyle \therefore P(\text{a spade})=\frac{10}{40}=\frac{1}{4}.

\displaystyle \text{(ii) All }4\text{ aces remain among the }40\text{ cards.}
\displaystyle \therefore \text{Number of cards which are not aces}=40-4=36.
\displaystyle \therefore P(\text{not an ace})=\frac{36}{40}=\frac{9}{10}.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{From a pack of }52\text{ cards, all aces and all kings are removed. A card is drawn}
\displaystyle \text{at random from the remaining cards. Find the probability that the card so drawn is}
\displaystyle \text{(i) a face card \qquad (ii) a card of red colour.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Number of cards removed}=4+4=8.
\displaystyle \therefore \text{Number of cards remaining}=52-8=44.
\displaystyle \text{(i) After removing all kings, the remaining face cards are }4\text{ queens and }4\text{ jacks.}
\displaystyle \therefore \text{Number of face cards remaining}=4+4=8.
\displaystyle \therefore P(\text{a face card})=\frac{8}{44}=\frac{2}{11}.

\displaystyle \text{(ii) Of the }26\text{ red cards, }2\text{ red aces and }2\text{ red kings are removed.}
\displaystyle \therefore \text{Number of red cards remaining}=26-4=22.
\displaystyle \therefore P(\text{a card of red colour})=\frac{22}{44}=\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{The number of red balls in a bag is three more than the number of black}
\displaystyle \text{balls. If the probability of drawing a red ball at random from the given bag is }\frac{12}{23},
\displaystyle \text{find the total number of balls in the given bag.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of black balls}=x.
\displaystyle \therefore \text{Number of red balls}=x+3.
\displaystyle \text{Total number of balls}=x+(x+3)=2x+3.
\displaystyle \text{Given, }P(\text{a red ball})=\frac{12}{23}.
\displaystyle \therefore \frac{x+3}{2x+3}=\frac{12}{23}.
\displaystyle 23(x+3)=12(2x+3).
\displaystyle 23x+69=24x+36.
\displaystyle x=33.
\displaystyle \therefore \text{Number of red balls}=33+3=36.
\displaystyle \therefore \text{Total number of balls}=33+36=69.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{A bag contains cards which are numbered from }5\text{ to }100\text{ such that each card}
\displaystyle \text{bears a different number. A card is drawn at random. Find the probability that the number}
\displaystyle \text{on the card is (i) a perfect square \qquad (ii) a }2\text{-digit number.}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=100-5+1=96.
\displaystyle \text{(i) Perfect square numbers from }5\text{ to }100\text{ are }9,16,25,36,49,64,81,100.
\displaystyle \therefore \text{Favourable number of elementary events}=8.
\displaystyle \therefore P(\text{a perfect square})=\frac{8}{96}=\frac{1}{12}.

\displaystyle \text{(ii) The }2\text{-digit numbers are from }10\text{ to }99.
\displaystyle \therefore \text{Favourable number of elementary events}=99-10+1=90.
\displaystyle \therefore P(\text{a }2\text{-digit number})=\frac{90}{96}=\frac{15}{16}.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{A bag contains balls numbered }2\text{ to }91\text{ such that each ball bears a different}
\displaystyle \text{number. A ball is drawn at random from the bag. Find the probability that}
\displaystyle \text{(i) it bears a }2\text{-digit number}\qquad\text{(ii) it bears a multiple of }1.
\displaystyle \hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of elementary events}=91-2+1=90.
\displaystyle \text{(i) The }2\text{-digit numbers are from }10\text{ to }91.
\displaystyle \therefore \text{Favourable number of elementary events}=91-10+1=82.
\displaystyle \therefore P(\text{a }2\text{-digit number})=\frac{82}{90}=\frac{41}{45}.

\displaystyle \text{(ii) Every number is a multiple of }1.
\displaystyle \therefore \text{Favourable number of elementary events}=90.
\displaystyle \therefore P(\text{a multiple of }1)=\frac{90}{90}=1.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Two customers are visiting a particular shop in the same week (Monday to}
\displaystyle \text{Saturday). Each is equally likely to visit the shop on any one day as on another. What is the}
\displaystyle \text{probability that both will visit the shop on:}
\displaystyle \text{(i) the same day?}\qquad\text{(ii) different days?}\qquad\text{(iii) consecutive days?}
\displaystyle \text{Answer:}
\displaystyle \text{There are }6\text{ days from Monday to Saturday.}
\displaystyle \therefore \text{Total number of elementary events}=6\times6=36.
\displaystyle \text{(i) Both customers can visit on the same day in }6\text{ ways.}
\displaystyle \therefore P(\text{same day})=\frac{6}{36}=\frac{1}{6}.

\displaystyle \text{(ii) Number of ways in which they visit on different days}=36-6=30.
\displaystyle \therefore P(\text{different days})=\frac{30}{36}=\frac{5}{6}.

\displaystyle \text{(iii) The consecutive pairs of days are (Mon, Tue), (Tue, Wed), (Wed, Thu),}
\displaystyle \text{(Thu, Fri) and (Fri, Sat).}
\displaystyle \text{Each pair gives }2\text{ possible ordered outcomes for the two customers.}
\displaystyle \therefore \text{Favourable number of elementary events}=5\times2=10.
\displaystyle \therefore P(\text{consecutive days})=\frac{10}{36}=\frac{5}{18}.
\displaystyle \\

 

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