\displaystyle \textbf{Question 1. }\text{The number of red balls in a bag is }10\text{ more than the number of black balls.}
\displaystyle \text{If the probability of drawing a red ball at random from this bag is }\frac{3}{5},\text{ then the total}
\displaystyle \text{number of balls in the bag is:} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) }50 \qquad \text{(b) }60 \qquad \text{(c) }80 \qquad \text{(d) }40
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{No. of black balls}=x
\displaystyle \text{No. of red balls}=x+10
\displaystyle \text{A.T.Q.}\ P(\text{Red ball})=\frac{x+10}{2x+10}=\frac{3}{5}
\displaystyle \Rightarrow 5x+50=6x+30
\displaystyle \Rightarrow 20=x
\displaystyle \text{Total no. of balls}=2\times20+10=50
\\

\displaystyle \textbf{Question 2. }\text{A pair of dice is thrown. The probability that sum of numbers appearing}
\displaystyle \text{on top faces is at most }10\text{ is:} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{(a) }\frac{1}{11} \qquad \text{(b) }\frac{10}{11} \qquad \text{(c) }\frac{5}{6} \qquad \text{(d) }\frac{11}{12}
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle P(E)=1-\frac{3}{36}\qquad [(5,6),(6,5),(6,6)]
\displaystyle =1-\frac{1}{12}
\displaystyle =\frac{11}{12}
\\

\displaystyle \textbf{Question 3. }\text{Two friends Anil and Ashraf were born in the December month}
\displaystyle \text{in the year 2010. Find the probability that:}
\displaystyle \text{(i) they share same date of birth.}
\displaystyle \text{(ii) they have different dates of birth.} \hspace{0.2cm}\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \text{Two friends Anil and Ashraf were born in december}
\displaystyle \text{month in year }2010.
\displaystyle \text{(i) Let }E\text{ be event that they share same date of birth}
\displaystyle \text{i.e., one day out of }31\text{ days. Then by definition}
\displaystyle P(E)=\frac{\text{No. of favourable outcomes}}{\text{Total no. of outcomes}}
\displaystyle \Rightarrow P(E)=\frac{1}{31}
\displaystyle \text{(ii) Let }\bar{E}\text{ be event that they have different dates of}
\displaystyle \text{birth.}
\displaystyle \text{As, }P(E)+P(\bar{E})=1
\displaystyle \Rightarrow P(\bar{E})=1-P(E)
\displaystyle \Rightarrow 1-\frac{1}{31}=\frac{30}{31}
\\

\displaystyle \textbf{Question 4. }\text{If the probability of a player winning a game is }0.79,\text{ then the probability}
\displaystyle \text{of his losing the same game is:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }1.79 \qquad \text{(b) }0.31 \qquad \text{(c) }0.21\% \qquad \text{(d) }0.21
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle P(\text{winning a game})=0.79
\displaystyle \text{Now, }P(\text{losing the game})=1-P(\text{winning the game})
\displaystyle =1-0.79=0.21
\\

\displaystyle \textbf{Question 5. }\text{From the data }1,4,7,9,16,21,25,\text{ if all the even numbers are removed,}
\displaystyle \text{then the probability of getting at random a prime number from the remaining is:} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{(a) }\frac{2}{5} \qquad \text{(b) }\frac{1}{5} \qquad \text{(c) }\frac{1}{7} \qquad \text{(d) }\frac{2}{7}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b)}
\displaystyle \text{The even numbers in the given data are }4\text{ and }16.
\displaystyle \text{Now, when even numbers are removed, then}
\displaystyle \text{possible outcomes will be }1,7,9,21,25.
\displaystyle \therefore \text{Total number of possible outcomes}=5
\displaystyle \text{Now, }7\text{ is the only prime number in the above data.}
\displaystyle \therefore \text{Number of favourable outcomes}=1
\displaystyle \therefore \text{Required probability}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}=\frac{1}{5}
\\

\displaystyle \textbf{Question 6. }\text{In a pack of }52\text{ playing cards one card is lost. From the remaining cards,}
\displaystyle \text{a card is drawn at random. Find the probability that the drawn card is queen of heart,}
\displaystyle \text{if the lost card is a black card.} \hspace{0.2cm}\text{[CBSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{When one card is lost, then }(52-1)\text{ i.e. }51\text{ cards are}
\displaystyle \text{left in the deck.}
\displaystyle \text{Total number of possible outcomes}=51
\displaystyle \text{Also, the lost card is a black card. Then, queen of}
\displaystyle \text{heart will also be left in the deck.}
\displaystyle \text{So, number of favourable outcomes}=1
\displaystyle \text{Required probability}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}=\frac{1}{51}
\\

\displaystyle \textbf{Question 7. }\text{Two dice are thrown together. The probability of getting the difference}
\displaystyle \text{of numbers on their upper faces equals to }3\text{ is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{9} \qquad \text{(b) }\frac{2}{9} \qquad \text{(c) }\frac{1}{6} \qquad \text{(d) }\frac{1}{12}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{Here, total number of outcomes}=36
\displaystyle \text{Let us define the event }A\text{ as,}
\displaystyle A=\text{Difference of number on their faces equals }3
\displaystyle \text{Favourable outcomes are }(1,4),(2,5),(3,6),(4,1),(5,2)\text{ and }(6,3)
\displaystyle P(A)=\frac{6}{36}=\frac{1}{6}
\\

\displaystyle \textbf{Question 8. }\text{A card is drawn at random from a well-shuffled pack of }52\text{ cards.}
\displaystyle \text{The probability that the card drawn is not an ace is:} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{13} \qquad \text{(b) }\frac{9}{13} \qquad \text{(c) }\frac{4}{13} \qquad \text{(d) }\frac{12}{13}
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{Total number of outcomes}=52
\displaystyle \text{Number of non ace cards}=48
\displaystyle \therefore \text{Probability that the card drawn is not an ace}
\displaystyle =\frac{48}{52}=\frac{12}{13}
\\

\displaystyle \textbf{Question 9. }\text{In a family of two children, the probability of having at least one girl is}
\displaystyle \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{(a) }\frac{1}{2} \qquad \text{(b) }\frac{2}{5} \qquad \text{(c) }\frac{3}{4} \qquad \text{(d) }\frac{1}{4}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{For a family of two children, the possible outcomes are: }
\displaystyle BB,\ BG,\ GB,\ GG
\displaystyle \text{where }G\rightarrow\text{girl; }B\rightarrow\text{boy}
\displaystyle \therefore \text{Total number of possible outcomes}=4
\displaystyle \text{Let }E\text{ be the event such that the family is having at least one girl.}
\displaystyle \text{So, outcomes favourable to event }E\text{ are:}
\displaystyle BG,\ GB,\ GG
\displaystyle \therefore \text{No. of favourable outcomes}=3
\displaystyle \therefore \text{Required probability}
\displaystyle =\frac{\text{No. of favourable outcomes}}{\text{Total possible outcomes}}=\frac{3}{4}
\\

\displaystyle \textbf{Question 10. }\text{If a letter of English alphabet is chosen at random, then the probability of}
\displaystyle \text{this letter to be a consonant is:} \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{(a) }\frac{5}{26} \qquad \text{(b) }\frac{21}{26} \qquad \text{(c) }\frac{10}{13} \qquad \text{(d) }\frac{11}{13}
\displaystyle \text{Answer:}
\displaystyle \textbf{(b)}
\displaystyle \text{In English alphabets we have }21\text{ consonants and }5
\displaystyle \text{vowels.}
\displaystyle \therefore \text{Probability of chosen letter is a consonant}=\frac{21}{26}
\\

\displaystyle \textbf{Question 11. }\text{A number is chosen from the numbers }1,2,3\text{ and denoted as }x\text{ and a}
\displaystyle \text{number is chosen from the numbers }1,4,9\text{ and denoted as }y.
\displaystyle \text{Then }P(y<x<9)\text{ is:}  \hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{(a) }\frac{1}{9} \qquad \text{(b) }\frac{3}{9} \qquad \text{(c) }\frac{5}{9} \qquad \text{(d) }\frac{7}{9}
\displaystyle \text{Answer:}
\displaystyle \textbf{(c)}
\displaystyle \text{The possible values of the product }xy\text{ are:}
\displaystyle 1,4,9,2,8,18,3,12,27
\displaystyle \therefore \text{Total number of possible outcomes}=9
\displaystyle \text{Let }E\text{ be the event such that the product }(xy)\text{ is less}
\displaystyle \text{than }9.
\displaystyle \text{So, outcomes favourable to event }E\text{ are:}
\displaystyle 1,2,3,4,8
\displaystyle \therefore \text{No. of favourable outcomes}=5
\displaystyle \therefore \text{Required probability}
\displaystyle =\frac{\text{No. of favourable outcomes}}{\text{Total possible outcomes}}=\frac{5}{9}
\\

\displaystyle \textbf{Question 12. }\text{If a fair coin is tossed twice, find the probability of getting} \\ \text{ `atmost one head'.} \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Possible outcomes are }HH,\ HT,\ TH,\ TT.
\displaystyle \text{Let }A\text{ be the event of getting at most one head.}
\displaystyle \text{Favourable outcomes of }A\text{ are }TT,\ TH,\ HT.
\displaystyle \therefore P(A)=\frac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}}=\frac{3}{4}
\\

\displaystyle \textbf{Question 13. }\text{Assertion (A): The probability that a leap year has }53\text{ Sundays is }\frac{2}{7}.
\displaystyle \text{Reason (R): The probability that a non-leap year has }53\text{ Sundays is }\frac{5}{7}. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{A leap year has }366\text{ days}=52\text{ weeks}+2\text{ days.}
\displaystyle \text{The two extra days can be }(Sun,Mon),(Mon,Tue),\ldots,(Sat,Sun).
\displaystyle \therefore \text{Probability of }53\text{ Sundays}=\frac{2}{7}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{A non-leap year has }365\text{ days}=52\text{ weeks}+1\text{ day.}
\displaystyle \therefore \text{Probability of }53\text{ Sundays}=\frac{1}{7}\ne\frac{5}{7}
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 14. }\text{Assertion (A): A fair die is thrown once. The probability of getting a prime}
\displaystyle \text{number is }\frac{1}{2}. \hspace{0.2cm}\text{[CBSE 2023]}
\displaystyle \text{Reason (R): A natural number is a prime number if it has only two factors.}
\displaystyle \text{(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct} \\ \text{explanation of Assertion (A).}
\displaystyle \text{(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct} \\ \text{explanation of Assertion (A).}
\displaystyle \text{(c) Assertion (A) is true but Reason (R) is false.}
\displaystyle \text{(d) Assertion (A) is false but Reason (R) is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Prime numbers on a die are }2,3,5.
\displaystyle \therefore \text{Required probability}=\frac{3}{6}=\frac{1}{2}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{A natural number having exactly two factors is called a prime number.}
\displaystyle \therefore \text{Reason (R) is true.}
\displaystyle \text{However, Reason (R) does not explain why the probability is }\frac{1}{2}.
\displaystyle \therefore \text{Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct}
\displaystyle \text{explanation of Assertion (A).}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 15. }\text{Assertion (A): Two players Sania and Ashnam play a tennis match. The}
\displaystyle \text{probability of Sania winning the match is }0.79\text{ and that of Ashnam winning the}
\displaystyle \text{match is }0.21.\hspace{0.2cm}\text{[CBSE 2023(C)]}
\displaystyle \text{Reason (R): The sum of probabilities of two complementary events is }1.
\displaystyle \text{Answer:}
\displaystyle 0.79+0.21=1
\displaystyle \text{Since winning by Sania and winning by Ashnam are complementary events, their probabilities add to }1.
\displaystyle \therefore \text{Both Assertion (A) and Reason (R) are true and Reason (R) is the correct}
\displaystyle \text{explanation of Assertion (A).}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 16. }\text{For an event }E,\ P(E)+P(\bar{E})=x,\text{ then the value of }
\displaystyle x^{3}-3\text{ is}  \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }-2 \qquad \text{(b) }2 \qquad \text{(c) }1 \qquad \text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{As, }P(E)+P(\bar{E})=1,\text{ so }x=1.
\displaystyle \text{Now, }x^{3}-3=(1)^{3}-3=-2
\\

\displaystyle \textbf{Question 17. }\text{The probability that the drawn card from a pack of }52\text{ cards is}
\displaystyle \text{neither an ace nor a spade is} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }\frac{9}{13} \qquad \text{(b) }\frac{35}{52} \qquad \text{(c) }\frac{10}{13} \qquad \text{(d) }\frac{19}{26}
\displaystyle \text{Answer:}
\displaystyle \textbf{(a)}
\displaystyle \text{Total number of possible outcomes}=52
\displaystyle \text{Number of favourable outcomes when drawn card is}
\displaystyle \text{neither an ace nor a spade}=36
\displaystyle \text{Required probability}
\displaystyle =\frac{\text{No. of favourable outcomes}}{\text{Total possible outcomes}}=\frac{36}{52}=\frac{9}{13}
\\

\displaystyle \textbf{Question 18. }\text{Which of the following cannot be the probability of an event?} \hspace{0.2cm}\text{[CBSE 2021]}
\displaystyle \text{(a) }0.01 \qquad \text{(b) }3\% \qquad \text{(c) }\frac{16}{17} \qquad \text{(d) }\frac{17}{16}
\displaystyle \text{Answer:}
\displaystyle \textbf{(d)}
\displaystyle \text{As for an event }E,\ 0\leq P(E)\leq1
\displaystyle \text{Now, }\frac{17}{16}>1,\text{ so it can't be the probability of an event.}
\\

\displaystyle \textbf{Question 19. }\text{A die is thrown once. What is the probability of getting a number} \\ \text{less than }3? \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes}=6
\displaystyle \text{Favourable outcomes}=2\ (\text{i.e. }1,2)
\displaystyle \therefore \text{Probability of getting a number less than }3=\frac{2}{6}=\frac{1}{3}
\\

\displaystyle \textbf{Question 20. }\text{If the probability of winning a game is }0.07,\text{ what is the} \\ \text{probability of losing it?} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ represents winning a game}
\displaystyle P(A)=0.07
\displaystyle P(\text{not }A)=P(\bar{A})
\displaystyle =1-P(A)=1-0.07
\displaystyle =0.93
\\

\displaystyle \textbf{Question 21. }\text{If a number }x\text{ is chosen at random from the numbers }-3,-2,-1,0,1,2,3.
\displaystyle \text{What is probability that }x^{2}\leq4? \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{If }x^{2}\leq4,\text{ then }x\text{ can take values }-2,-1,0,1,2.
\displaystyle \therefore \text{Probability of selecting }x\text{ such that }x^{2}\leq4=\frac{5}{7}
\\

\displaystyle \textbf{Question 22. }\text{A die is thrown once. What is the probability of getting an even} \\ \text{prime number?} \hspace{0.2cm}\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Total outcomes for a throw of a die}=6
\displaystyle \text{i.e. numbers }1,2,3,4,5,6
\displaystyle \text{Favourable outcomes for getting an even prime number}
\displaystyle =1\text{ i.e. number }2
\displaystyle \therefore \text{Probability of getting an even prime number}=\frac{1}{6}
\\

\displaystyle \textbf{Question 23. }\text{A game consists of tossing a coin }3\text{ times and noting the }
\displaystyle \text{outcome each time. If getting the same result in all the tosses is a success, }
\displaystyle \text{find the  probability of losing the game.} \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{A coin is tossed }3\text{ times. The possible outcomes are:}
\displaystyle \{HHH,\ HTH,\ THH,\ HHT,\ TTT,\ THT,\ HTT,\ TTH\}
\displaystyle \text{Total number of possible outcomes}=8
\displaystyle \text{Favourable outcomes of getting same result in all the}
\displaystyle \text{tosses are }HHH\text{ and }TTT
\displaystyle \text{Number of favourable outcomes}=2
\displaystyle \text{Probability of winning}=\frac{2}{8}=\frac{1}{4}
\displaystyle \text{So, probability of losing}=1-\frac{1}{4}=\frac{3}{4}
\\

\displaystyle \textbf{Question 24. }\text{A die is thrown once. Find the probability of getting a number which}
\displaystyle \text{(i) is a prime number.}
\displaystyle \text{(ii) lies between }2\text{ and }6. \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{When a die is thrown, then possible outcomes are}
\displaystyle \{1,2,3,4,5,6\}
\displaystyle \text{Total number of possible outcomes}=6
\displaystyle \text{(i) Prime numbers are }2,3\text{ and }5.
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \text{Required probability}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{(ii) Number between }2\text{ and }6\text{ are }3,4\text{ and }5.
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \text{Required probability}=\frac{3}{6}=\frac{1}{2}
\\

\displaystyle \textbf{Question 25. }\text{The probability of selecting a blue marble at random from a jar }
\displaystyle \text{that contains only blue, black and green marbles is }\frac{1}{5}. \text{The probability of selecting a }
\displaystyle \text{black marble at random from the same jar is }\frac{1}{4}. \text{If the jar contains }11\text{ green marbles, }
\displaystyle \text{find the total number of marbles in the jar.} \hspace{0.2cm}\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total number of marbles in the jar}=x
\displaystyle \text{Probability of selecting a blue marble}+\text{Probability of}
\displaystyle \text{selecting a black marble}+\text{Probability of selecting}
\displaystyle \text{a green marble}=1
\displaystyle \frac{1}{5}+\frac{1}{4}+\frac{11}{x}=1
\displaystyle \Rightarrow \frac{11}{x}=1-\frac{1}{5}-\frac{1}{4}
\displaystyle \Rightarrow \frac{11}{x}=\frac{20-4-5}{20}
\displaystyle \Rightarrow \frac{11}{x}=\frac{11}{20}\Rightarrow x=20
\\

\displaystyle \textbf{Question 26. }\text{Two different dice are tossed together. Find the probability:} \hspace{0.2cm}\text{[CBSE 2018]}
\displaystyle \text{(i) of getting a doublet}
\displaystyle \text{(ii) of getting a sum }10,\text{ of the numbers on the two dice.}
\displaystyle \text{Answer:}
\displaystyle \text{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 \text{Total elementary events}=36
\displaystyle \text{(i) Outcomes of doublet}
\displaystyle =\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}\text{ i.e. }6
\displaystyle P(\text{getting a doublet})=\frac{6}{36}=\frac{1}{6}
\displaystyle \text{(ii) Outcomes of getting a sum }10
\displaystyle =\{(4,6),(5,5),(6,4)\}\text{ i.e. }3
\displaystyle P(\text{getting a sum }10)=\frac{3}{36}=\frac{1}{12}
\\

\displaystyle \textbf{Question 27. }\text{An integer is chosen at random between }1\text{ and }100. \\ \text{Find the probability that it is:}
\displaystyle \text{(i) divisible by }8.
\displaystyle \text{(ii) not divisible by }8. \hspace{0.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) The integers divisible by }8\text{ between }1\text{ and }100
\displaystyle \text{are }8,16,24,32,40,48,56,64,72,80,88,96,
\displaystyle \text{i.e. }12
\displaystyle \text{Total outcomes}=98
\displaystyle P(\text{divisible by }8)=\frac{12}{98}=\frac{6}{49}
\displaystyle \text{(ii) }P(\text{not divisible by }8)
\displaystyle =\frac{98-12}{98}=\frac{86}{98}=\frac{43}{49}
\\

\displaystyle \textbf{Question 28. }\text{The King, Queen and Jack of clubs are removed from a pack of }52\text{ cards}
\displaystyle \text{and then the remaining cards are well shuffled. A card is selected from the remaining cards.}
\displaystyle \text{Find the probability of getting a} \hspace{0.2cm}\text{[CBSE 2018(C)]}
\displaystyle \text{(i) of spade}
\displaystyle \text{(ii) of black king}
\displaystyle \text{(iii) of club}
\displaystyle \text{(iv) of jack}
\displaystyle \text{Answer:}
\displaystyle \text{King, Queen and Jack of clubs are removed from a}
\displaystyle \text{pack of }52\text{ cards.}
\displaystyle \text{Remaining cards}=52-3=49
\displaystyle =\text{Total number of possible outcomes.}
\displaystyle \text{(i) Number of spades}=13
\displaystyle P(\text{spade})=\frac{13}{49}
\displaystyle \text{(ii) Number of black king}=1
\displaystyle P(\text{black king})=\frac{1}{49}
\displaystyle \text{(iii) Number of clubs}=10
\displaystyle P(\text{club})=\frac{10}{49}
\displaystyle \text{(iv) Number of Jacks}=3
\displaystyle P(\text{Jack})=\frac{3}{49}
\\

\displaystyle \textbf{Question 29. }\text{The probability of selecting a rotten apple randomly from a heap of }
\displaystyle 900\text{ apples } \text{is }0.18. \text{What is the number of rotten apples in the heap?} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Total apples in the heap}=900
\displaystyle \therefore \text{Total number of elementary events}=900
\displaystyle \text{One rotten apple is randomly selected from the heap.}
\displaystyle P(\text{a rotten apple})=\frac{\text{Favourable events}}{\text{Total elementary events}}
\displaystyle 0.18=\frac{\text{Favourable events}}{900}
\displaystyle \Rightarrow \text{Favourable events}=0.18\times900=162
\displaystyle \text{Hence, there are }162\text{ rotten apples in the heap.}
\\

\displaystyle \textbf{Question 30. }\text{A bag contains }3\text{ red and }5\text{ black balls. A ball is drawn at random}
\displaystyle \text{from the bag. What is the probability that the drawn ball is not red?} \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Number of red balls}=3
\displaystyle \text{Number of black balls}=5
\displaystyle \text{Total balls}=8
\displaystyle \text{Probability that the ball drawn is not red}=\frac{5}{8}
\\

\displaystyle \textbf{Question 31. }\text{Two different dice are thrown together. Find the probability that} \\ \text{the numbers obtained}
\displaystyle \text{(i) have a sum less than }7
\displaystyle \text{(ii) have a product less than }16
\displaystyle \text{(iii) is a doublet of odd number.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Elementary events associated with the random}
\displaystyle \text{experiment of throwing two dice 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 \text{Total number of elementary events}=6\times6=36
\displaystyle \text{(i) Let }A\text{ be the event of getting a sum less than}
\displaystyle 7\text{ in the numbers obtained on the two dice.}
\displaystyle \therefore \text{Elementary events favourable to event }A
\displaystyle \text{ are:}
\displaystyle (1,1),(2,1),(3,1),(4,1),(5,1),(1,2),(1,3),
\displaystyle (1,4),(1,5),(2,2),(2,3),(2,4),(3,2),(3,3),
\displaystyle (4,2)
\displaystyle \text{Favourable number of elementary events}=15
\displaystyle \text{Hence, required probability}=\frac{15}{36}=\frac{5}{12}
\displaystyle \text{(ii) Let }B\text{ be the event of getting a product less}
\displaystyle \text{than }16.
\displaystyle \therefore \text{Elementary events favourable to event }B
\displaystyle \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),(4,1),(4,2),
\displaystyle (4,3),(5,1),(5,2),(5,3),(6,1),(6,2)
\displaystyle \Rightarrow \text{Favourable number of elementary events}=25
\displaystyle \text{Hence, required probability}=\frac{25}{36}
\displaystyle \text{(iii) Let }C\text{ be the event of getting a doublet of odd}
\displaystyle \text{number.}
\displaystyle \text{Elementary events favourable to the event }C
\displaystyle \text{ are: }(1,1),(3,3),(5,5)
\displaystyle \therefore \text{Favourable number of elementary events}=3
\displaystyle \therefore \text{Required probability}=\frac{3}{36}=\frac{1}{12}
\\

\displaystyle \textbf{Question 32. }\text{A bag contains }15\text{ white and some black balls. If the }
\displaystyle \text{probability of drawing a black ball from the bag is thrice that of drawing a  }
\displaystyle \text{white ball, find the number of black balls in the bag.} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of black balls}=x
\displaystyle \text{Number of white balls}=15
\displaystyle \text{Total number of balls}=x+15
\displaystyle \therefore \text{Total number of possible outcomes}=x+15
\displaystyle P(\text{a black ball})=\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}
\displaystyle =\frac{\text{Number of black balls}}{\text{Total number of balls}}=\frac{x}{x+15}
\displaystyle \text{Similarly, }P(\text{a white ball})=\frac{15}{x+15}
\displaystyle \text{A.T.Q., }P(\text{a black ball})=3\times P(\text{a white ball})
\displaystyle \Rightarrow \frac{x}{x+15}=3\times\frac{15}{x+15}
\displaystyle \Rightarrow x=45
\displaystyle \therefore \text{Number of black balls}=45
\\

\displaystyle \textbf{Question 33. }\text{A lot consists of }144\text{ ball pens of which }20\text{ are defective. The customer}
\displaystyle \text{will buy a ball pen if it is good, but will not buy a defective ball pen. The shopkeeper}
\displaystyle \text{draws one pen at random from the lot and gives it to the consumer. What is the probability that}
\displaystyle \text{(i) customer will buy the ball pen}
\displaystyle \text{(ii) customer will not buy the ball pen} \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{Answer:}
\displaystyle \text{Total ball pens}=144
\displaystyle \text{Defective ball pens}=20
\displaystyle \text{Good ball pens}=144-20=124
\displaystyle \text{(i) Probability that the customer will buy the ball}
\displaystyle \text{pen}=\frac{124}{144}=\frac{31}{36}
\displaystyle \text{(ii) Probability that the customer will not buy the}
\displaystyle \text{ball pen}=1-\frac{31}{36}=\frac{5}{36}
\\

\displaystyle \textbf{Question 34. }\text{Peter throws two different dice together and finds the product }
\displaystyle \text{of the two numbers obtained. Rina throws a die and squares the number obtained. }?
\displaystyle \text{Who has the better chance to get the number }25 \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Peter throws two different dice and Rina throws only one die.}
\displaystyle \text{Let us calculate the probability in each case. When a pair of dice is thrown, there are }
\displaystyle 36\text{ elementary events} \text{ as given below:}
\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 \text{The product of two numbers on the two dice will be }25\text{ if both the dice show number }5.
\displaystyle \text{So, there is only one elementary event favourable to getting }25\text{ (a '}5\text{' on each die).}
\displaystyle \therefore P_{1}(\text{Peter getting }25)=\frac{1}{36}
\displaystyle \text{Rina throws only one die on which she can get one of the six numbers }1,2,3,4,5,6.
\displaystyle \text{If she gets number }5\text{ on the upper face of the die thrown, then the square of the number is }25.
\displaystyle \therefore P_{2}(\text{Rina getting a number whose square is }25)=\frac{1}{6}
\displaystyle \text{Clearly }P_{2}>P_{1}.
\displaystyle \text{So, Rina has better chance of getting the number }25.
\\

\displaystyle \textbf{Question 35. }\text{Two different dice are thrown together. Find the probability} \\ \text{that the numbers obtained} \hspace{0.2cm}\text{[CBSE 2017]}
\displaystyle \text{(i) even sum}
\displaystyle \text{(ii) even product}
\displaystyle \text{Answer:}
\displaystyle \text{When two different dice are thrown together, total number of possible outcomes }
\displaystyle \text{is }36\text{ as shown below.}
\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 \text{(i) Let }E_{1}\text{ be the event of getting an even sum.}
\displaystyle \therefore \text{Outcomes favourable to the event }E_{1}\text{ are:}
\displaystyle (1,1),(1,3),(1,5),(2,2),(2,4),(2,6),
\displaystyle (3,1),(3,3),(3,5),(4,2),(4,4),(4,6),
\displaystyle (5,1),(5,3),(5,5),(6,2),(6,4),(6,6)
\displaystyle \text{Number of favourable outcomes to the event }E_{1}=18
\displaystyle \therefore P(E_{1})=\frac{18}{36}=\frac{1}{2}
\displaystyle \text{(ii) Let }E_{2}\text{ be the event of getting even product.}
\displaystyle \therefore \text{Outcomes favourable to the event }E_{2}\text{ are}
\displaystyle (1,2),(1,4),(1,6),(2,1),(2,2),(2,3),
\displaystyle (2,4),(2,5),(2,6),(3,2),(3,4),(3,6),
\displaystyle (4,1),(4,2),(4,3),(4,4),(4,5),(4,6),
\displaystyle (5,2),(5,4),(5,6),(6,1),(6,2),(6,3),
\displaystyle (6,4),(6,5),(6,6)\text{ i.e. }27.
\displaystyle \therefore P(E_{2})=\frac{27}{36}=\frac{3}{4}
\\

\displaystyle \textbf{Question 36. }\text{A box contains cards, number from }1\text{ to }90. \text{A card is drawn at random}
\displaystyle \text{from the box. Find the probability that the selected card bears a} \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{(i) two digit number.}
\displaystyle \text{(ii) perfect square number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total cards}=90
\displaystyle \text{Two digit numbers are }10\text{ to }90.
\displaystyle \text{Number of cards bearing two digit number}=81
\displaystyle \text{(i) Probability of getting a card bearing a }2\text{-digit}
\displaystyle \text{number}=\frac{81}{90}=\frac{9}{10}
\displaystyle \text{(ii) Perfect squares from }1\text{ to }90\text{ are }1,4,9,16,
\displaystyle 25,36,49,64,81.
\displaystyle \text{Number of perfect square numbers}=9
\displaystyle \text{Probability of getting a card bearing a perfect}
\displaystyle \text{square number}=\frac{9}{90}=\frac{1}{10}
\\

\displaystyle \textbf{Question 37. }\text{From a pack of }52\text{ playing cards, Jacks and Kings of red colour and Queens}
\displaystyle \text{and Aces of black colour are removed. The remaining cards are mixed and a card is drawn}
\displaystyle \text{at random. Find the probability that the drawn card is} \hspace{0.2cm}\text{[CBSE 2017(C)]}
\displaystyle \text{(i) a black Queen}
\displaystyle \text{(ii) a card of red colour}
\displaystyle \text{(iii) a Jack of black colour}
\displaystyle \text{(iv) a face card}
\displaystyle \text{Answer:}
\displaystyle \text{Number of playing cards}=52
\displaystyle \text{Jacks and Kings of red colour and Queens and Aces of black colour are removed, therefore,}
\displaystyle \text{remaining cards are }44.
\displaystyle \text{(i) Number of black queens left}=0
\displaystyle \text{Probability of a black queen}=\frac{0}{44}=0
\displaystyle \text{(ii) Remaining red colour cards}=22
\displaystyle \text{Probability of getting a red colour card}=\frac{22}{44}=\frac{1}{2}
\displaystyle \text{(iii) Number of Jack of black colour}=2
\displaystyle \text{Probability of getting a Jack of black colour}=\frac{2}{44}=\frac{1}{22}
\displaystyle \text{(iv) Remaining face cards}=6
\displaystyle \text{Probability of getting a face card}=\frac{6}{44}=\frac{3}{22}
\\

\displaystyle \textbf{Question 38. }\text{Two different dice are thrown together. Find the probability of:} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{(i) getting a number greater than }3\text{ on each die.}
\displaystyle \text{(ii) getting a total of }6\text{ or }7\text{ of the numbers on two dice.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) When two dice are thrown together total}
\displaystyle \text{possible outcomes}=6\times6=36
\displaystyle \text{Favourable outcomes when both dice have}
\displaystyle \text{number more than }3\text{ are }(4,4),(4,5),
\displaystyle (4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)
\displaystyle \text{i.e. }9\text{ outcomes.}
\displaystyle P(\text{a number greater than }3\text{ on each dice})
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}}
\displaystyle =\frac{9}{36}=\frac{1}{4}
\displaystyle \text{(ii) Favourable outcomes when sum of the numbers}
\displaystyle \text{appearing on the dice is }6\text{ or }7\text{ are, }(1,5),(1,6),
\displaystyle (2,4),(2,5),(3,3),(3,4),(4,2),(4,3),(5,1),
\displaystyle (5,2),(6,1),\text{ i.e. }11\text{ outcomes.}
\displaystyle P(\text{a total of }6\text{ or }7)=\frac{11}{36}
\\

\displaystyle \textbf{Question 39. }\text{A box consists of }100\text{ shirts of which }88\text{ are good, }8\text{ have minor defects}
\displaystyle \text{and }4\text{ have major defects. Ramesh, a shopkeeper will buy only those shirts which are good}
\displaystyle \text{but `Kewal' another shopkeeper will not buy shirts with major defects. A shirt is taken}
\displaystyle \text{out of the box at random. What is the probability} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{(i) Ramesh will buy the selected shirt?}
\displaystyle \text{(ii) `Kewal' will buy the selected shirt?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) When one shirt is taken out of a box containing}
\displaystyle 100\text{ shirts, then number of total possible}
\displaystyle \text{outcomes}=100
\displaystyle \therefore \text{Number of favourable outcomes}
\displaystyle =\text{number of good shirts}=88
\displaystyle P(\text{Ramesh buys a shirt})
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}}
\displaystyle =\frac{88}{100}=\frac{22}{25}
\displaystyle \text{(ii) Kewal will buy a shirt if the shirt is not having}
\displaystyle \text{major defect.}
\displaystyle \text{Number of favourable outcomes}=\text{Number of}
\displaystyle \text{shirts without major defect}=100-4=96
\displaystyle P(\text{Kewal buys a shirt})=\frac{96}{100}=\frac{24}{25}
\\

\displaystyle \textbf{Question 40. }\text{Three different coins are tossed together. Find the probability} \\ \text{of getting} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{(i) exactly two heads}
\displaystyle \text{(ii) at least two heads}
\displaystyle \text{(iii) at least two tails.}
\displaystyle \text{Answer:}
\displaystyle \text{Possible outcomes when three coins are tossed:}
\displaystyle HHH,\ HHT,\ HTT,\ TTT,\ THH,\ TTH,\ HTH,\ THT
\displaystyle \text{Total number of possible outcomes}=8
\displaystyle \text{(i) For exactly two heads favourable outcomes are}
\displaystyle HHT,\ HTH\text{ and }THH.
\displaystyle P(\text{exactly two heads})=\frac{3}{8}
\displaystyle \text{(ii) In case of at least two heads, outcomes are }HHH,
\displaystyle HHT,\ THH\text{ and }HTH.
\displaystyle P(\text{at least two heads})=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{(iii) In case of at least two tails, outcomes are }TTH,
\displaystyle THT,\ HTT\text{ and }TTT.
\displaystyle P(\text{at least two tails})=\frac{4}{8}=\frac{1}{2}
\\

\displaystyle \textbf{Question 41. }\text{There are }100\text{ cards in a bag on which numbers from }1\text{ to }100\text{ are written.}
\displaystyle \text{A card is taken out from the bag at random. Find the probability that the number}
\displaystyle \text{on the selected card:} \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{(i) is divisible by }9\text{ and is a perfect square.}
\displaystyle \text{(ii) is a prime number greater than }80.
\displaystyle \text{Answer:}
\displaystyle \text{Total possible cases}=100
\displaystyle \text{(i) Favourable cases when number is a perfect}
\displaystyle \text{square and is divisible by }9\text{ are }9,36\text{ and }81.
\displaystyle \text{So, number of favourable cases}=3
\displaystyle \text{Required probability}
\displaystyle =\frac{\text{Number of favourable cases}}{\text{Total possible cases}}=\frac{3}{100}
\displaystyle \text{(ii) Favourable cases for the prime numbers greater}
\displaystyle \text{than }80\text{ are }83,89\text{ and }97.
\displaystyle \text{So, number of favourable cases}=3
\displaystyle \text{Required probability}
\displaystyle =\frac{\text{Number of favourable cases}}{\text{Total possible cases}}=\frac{3}{100}
\\

\displaystyle \textbf{Question 42. }\text{A game of chance consists of spinning an arrow on a circular board, divided into }8
\displaystyle \text{equal parts, which comes to rest pointing at one of the numbers }1,2,3,\ldots,8\text{ in the given}
\displaystyle \text{figure, which are equally likely outcomes. What is the probability that the arrow will point at}
\displaystyle \text{(i) an odd number}
\displaystyle \text{(ii) a number greater than }3
\displaystyle \text{(iii) a number less than }9. \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Total possible outcomes when the arrow points at one of the numbers are }8.
\displaystyle \text{Favourable outcomes when the required number is odd are }1,3,5,7,\text{ i.e. }4\text{ outcomes.}
\displaystyle \therefore P(\text{an odd number})=\frac{\text{Favourable outcomes}}{\text{Total possible outcomes}}=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{(ii) Favourable outcomes when the required number is more than }3\text{ are }4,5,6,7,8,\text{ i.e. }5\text{ outcomes.}
\displaystyle P(\text{a number is more than }3)=\frac{\text{Favourable outcomes}}{\text{Total possible outcomes}}=\frac{5}{8}
\displaystyle \text{(iii) Favourable outcomes when the required number is less than }9\text{ are }1,2,3,4,5,6,7,8,\text{ i.e. }8\text{ outcomes.}
\displaystyle P(\text{number is less than }9)=\frac{\text{Favourable outcomes}}{\text{Total possible outcomes}}=\frac{8}{8}=1
\\

\displaystyle \textbf{Question 43. }\text{Figure shown a disc on which a player spins an arrow twice. The fraction }\frac{a}{b}
\displaystyle \text{is formed, where }a\text{ is the number of sector on which arrow stops on the first spin and }b
\displaystyle \text{is the number of the sector on which the arrow stops on second spin. On each spin,}
\displaystyle \text{each sector has equal chance of selection by the arrow. Find the probability that }\frac{a}{b}>1.
\displaystyle \hspace{0.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{For }\frac{a}{b}>1,\text{ when, }a=1,\ b\text{ can not take any value}
\displaystyle \text{when, }a=2,\ b\text{ can take }1\text{ value i.e. }1
\displaystyle \text{when, }a=3,\ b\text{ can take }2\text{ values, i.e. }1\text{ and }2
\displaystyle \text{when, }a=4,\ b\text{ can take }3\text{ values, i.e. }1,2,3
\displaystyle \text{when, }a=5,\ b\text{ can take }4\text{ values, i.e. }1,2,3,4
\displaystyle \text{when, }a=6,\ b\text{ can take }5\text{ values, i.e. }1,2,3,4,5
\displaystyle \text{Total possible outcomes}=6\times6=36
\displaystyle \text{Favourable outcomes}=1+2+3+4+5=15
\displaystyle P\left(\frac{a}{b}>1\right)=\frac{\text{Number of favourable outcomes}}{\text{Number of total possible outcomes}}
\displaystyle =\frac{15}{36}=\frac{5}{12}
\\

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