\displaystyle \textbf{Question 1.}~\text{A die is thrown once. What is the probability that the}~\text{[ICSE 2009]}
\displaystyle (i)~\text{number is even?}
\displaystyle (ii)~\text{number is greater than }2?
\displaystyle \text{Answer:}
\displaystyle \text{When a die is thrown, then total number of outcomes}=6~\text{i.e. }1,2,3,4,5,6.
\displaystyle \text{(i) Even numbers on a die}=\{2,4,6\}
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \therefore~\text{Probability of getting an even number}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{(ii) Number greater than }2\text{ on a die}=\{3,4,5,6\}
\displaystyle \text{Number of favourable outcomes}=4
\displaystyle \therefore~\text{Probability of getting a number greater than }2
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{4}{6}=\frac{2}{3}

\displaystyle \textbf{Question 2.}~\text{Cards marked with numbers }1,2,3,4,\ldots,20\text{ are well shuffled} \\ \text{and a card is drawn at random. What is the probability that the number of} \\ \text{the card is}~\text{[ICSE 2010]}
\displaystyle (i)~\text{a prime number?}
\displaystyle (ii)~\text{divisible by }3?
\displaystyle (iii)~\text{a perfect square?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, cards marked with numbers }1,2,\ldots,20.
\displaystyle \text{Total number of cards}=20
\displaystyle \text{(i) Prime numbers}=\{2,3,5,7,11,13,17,19\}
\displaystyle \text{Number of favourable cards}=8
\displaystyle \therefore~\text{Probability of getting a card of prime number}
\displaystyle =\frac{\text{Number of favourable cards}}{\text{Total number of cards}}=\frac{8}{20}=\frac{2}{5}
\displaystyle \text{(ii) Numbers divisible by }3=\{3,6,9,12,15,18\}
\displaystyle \text{Number of favourable cards}=6
\displaystyle \therefore~\text{Probability of getting a card, whose number is divisible by }3
\displaystyle =\frac{\text{Number of favourable cards}}{\text{Total number of cards}}=\frac{6}{20}=\frac{3}{10}
\displaystyle \text{(iii) Perfect square numbers}=\{1,4,9,16\}
\displaystyle \text{Number of favourable cards}=4
\displaystyle \therefore~\text{Probability of getting a card of perfect square number}
\displaystyle =\frac{\text{Number of favourable cards}}{\text{Total number of cards}}=\frac{4}{20}=\frac{1}{5}

\displaystyle \textbf{Question 3.}~\text{A box contains some black balls and 30 white balls. If the probability} \\ \text{of drawing a black ball is two-fifths of the probability of drawing a white ball, then} \\ \text{find the number of black balls in the box.}~\text{[ICSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let number of black balls be }x.
\displaystyle \text{Given, number of white balls}=30
\displaystyle \text{So, total number of balls}=x+30
\displaystyle \text{Now, probability of drawing a white ball}
\displaystyle =\frac{\text{Number of white balls}}{\text{Total number of balls}}=\frac{30}{x+30}
\displaystyle \text{and probability of drawing a black ball}
\displaystyle =\frac{\text{Number of black balls}}{\text{Total number of balls}}=\frac{x}{x+30}
\displaystyle \text{According to the question,}
\displaystyle \text{Probability of drawing a black ball}=\frac{2}{5}\left[\text{probability of drawing a white ball}\right]
\displaystyle \therefore~\frac{x}{x+30}=\frac{2}{5}\left(\frac{30}{x+30}\right)\Rightarrow x=\frac{2}{5}\times 30=12
\displaystyle \text{Hence, the number of black balls in the box is }12.

\displaystyle \textbf{Question 4.}~\text{A die has 6 faces marked by the given numbers as shown below:}~\text{[ICSE 2014]}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline 1&2&3&-1&-2&-3\\\hline\end{array}
\displaystyle \text{The die is thrown once, what is the probability of getting}
\displaystyle (i)~\text{a positive integer?}\qquad (ii)~\text{an integer greater than }-3?
\displaystyle (iii)~\text{the smallest integer?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of sample points for given die}=6
\displaystyle \text{(i) Positive integers}=\{1,2,3\}
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \therefore~\text{Probability of getting a positive integer}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{(ii) Integers greater than }-3\text{ are }\{1,2,3,-1,-2\}.
\displaystyle \text{Number of favourable outcomes}=\text{Number of integers greater than }-3=5
\displaystyle \therefore~\text{Probability of getting an integer greater than }-3
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{5}{6}
\displaystyle \text{(iii) Smallest integer is }-3\text{ only.}
\displaystyle \text{Number of favourable outcomes}=1
\displaystyle \therefore~\text{Probability of getting a smallest integer}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{1}{6}

\displaystyle \textbf{Question 5.}~\text{A bag contains 5 white balls, 6 red balls and 9 green balls. A} \\ \text{ball is drawn at random from the bag. Find the probability that the ball} \\ \text{drawn is [ICSE 2015]}
\displaystyle (i)~\text{a green ball}
\displaystyle (ii)~\text{a white or a red ball}
\displaystyle (iii)~\text{neither a green ball nor a white ball.}
\displaystyle \text{Answer:}
\displaystyle \text{A bag containing }5\text{ white balls, }6\text{ red balls and }9\text{ green balls.}
\displaystyle \text{Total number of balls}=20
\displaystyle \text{(i) Probability of a green ball, }P(G)=\frac{9}{20}
\displaystyle \text{(ii) Probability of a white or red ball}=P(W\text{ or }R)=P(W)+P(R)
\displaystyle =\frac{5}{20}+\frac{6}{20}=\frac{11}{20}
\displaystyle \text{(iii) Probability of neither a green ball nor a white ball}
\displaystyle =\frac{20-9-5}{20}=\frac{6}{20}=\frac{3}{10}

\displaystyle \textbf{Question 6.}~\text{A game of numbers has cards marked with }11,12,13,\ldots,40.\text{ A} \\ \text{card is drawn at random. Find the probability that the number on the card} \\ \text{drawn is}~\text{[ICSE 2016]}
\displaystyle (i)~\text{a perfect square}\qquad (ii)~\text{divisible by }7.
\displaystyle \text{Answer:}
\displaystyle \text{Given, cards marked with numbers }11,12,13,\ldots,40.
\displaystyle \text{Total number of cards}=30
\displaystyle \text{(i) A perfect square number}=\{16,25,36\}
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \therefore~\text{Required probability}=\frac{3}{30}=\frac{1}{10}
\displaystyle \text{(ii) Number divisible by }7=\{14,21,28,35\}
\displaystyle \text{Number of favourable outcomes}=4
\displaystyle \therefore~\text{Required probability}=\frac{4}{30}=\frac{2}{15}

\displaystyle \textbf{Question 7.}~\text{From a pack of 52 playing cards, all cards, whose numbers are multiples} \\ \text{of 3, are removed. A card is now drawn at random. What is the probability that} \\ \text{the card drawn is}~\text{[ICSE 2011]}
\displaystyle (i)~\text{a face card (king, jack or queen)?}
\displaystyle (ii)~\text{an even numbered red card?}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards in a pack}=52
\displaystyle \text{Multiples of }3\text{ in each suit are }3,6,9.
\displaystyle \text{We know that in a pack of cards, there are four suits.}
\displaystyle \text{Number of removed cards}=3\times 4=12
\displaystyle \text{Total number of cards left}=52-12=40
\displaystyle \text{(i) Number of favourable cards}=3\times 4=12
\displaystyle \text{[since, face cards are }3\text{ in each suit and there are }4\text{ suits]}
\displaystyle \therefore~\text{Probability of getting a face card}
\displaystyle =\frac{\text{Number of favourable cards}}{\text{Total number of cards}}=\frac{12}{40}=\frac{3}{10}
\displaystyle \text{(ii) Even numbered red cards}=2\times \{2,4,8,10\}=2\times 4=8
\displaystyle \text{[since, there are two suits of red cards and }6\text{ is already removed]}
\displaystyle \text{Number of favourable cards}=8
\displaystyle \therefore~\text{Probability of getting an even numbered red card}
\displaystyle =\frac{\text{Number of favourable cards}}{\text{Total number of cards}}=\frac{8}{40}=\frac{1}{5}

\displaystyle \textbf{Question 8.}~\text{Sixteen cards are labelled as }a,b,c,\ldots,m,n,o,p.\text{ They are put} \\ \text{in a box and shuffled. A boy is asked to draw a card from the box. What} \\ \text{is the probability that the card drawn is}~\text{[ICSE 2017]}
\displaystyle (i)~\text{a vowel?}\qquad (ii)~\text{a consonant?}
\displaystyle (iii)~\text{none of the letters of the word }\text{median}?
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=16
\displaystyle \text{Clearly, the cards labelled }a,e,i,o\text{ are vowels and rest }12\text{ are consonant.}
\displaystyle \text{(i) There are }4\text{ ways of drawing a vowel card.}
\displaystyle \therefore~\text{Probability of drawing a vowel card}=\frac{4}{16}=\frac{1}{4}
\displaystyle \text{(ii) There are }12\text{ ways of drawing a consonant card.}
\displaystyle \therefore~\text{Probability of drawing a consonant card}=\frac{12}{16}=\frac{3}{4}
\displaystyle \text{(iii) Number of cards not labelled with letters of the word median}=16-6=10
\displaystyle \therefore~\text{Number of ways of drawing a card not labelled with letters of the word median}=10
\displaystyle \therefore~\text{Probability of drawing a card not labelled with letters of the median}=\frac{10}{16}=\frac{5}{8}

\displaystyle \textbf{Question 9.}~\text{A fair die is rolled. Find the probability of getting}~\text{[ICSE 2017]}
\displaystyle (i)~3\text{ on the face of the dice.}
\displaystyle (ii)~\text{an odd number on the face of the dice.}
\displaystyle (iii)~\text{a number greater than }1\text{ on the face of the dice.}
\displaystyle \text{Answer:}
\displaystyle \text{When a die is rolled.}
\displaystyle \text{Total number of outcomes}=6=\{1,2,3,4,5,6\}
\displaystyle \text{(i) Probability of getting }3\text{ on the face of dice}=\frac{1}{6}
\displaystyle \text{(ii) Favourable outcomes for an odd number on the face}=3=\{1,3,5\}
\displaystyle \therefore~\text{Probability of getting odd number}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{(iii) Favourable outcomes of a number greater than }1\text{ on the face}=5=\{2,3,4,5,6\}
\displaystyle \therefore~\text{Probability of getting a number greater than }1=\frac{5}{6}

\displaystyle \textbf{Question 10.}~\text{An integer is chosen at random from 1 to 50. Find the probability} \\ \text{that the number is}~\text{[ICSE 2017]}
\displaystyle (i)~\text{divisible by }5\qquad (ii)~\text{a perfect cube}\qquad (iii)~\text{a prime number}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of integers (from }1\text{ to }50)=50
\displaystyle \text{and integers divisible by }5=10=\{5,10,15,20,25,30,35,40,45,50\}
\displaystyle \text{and perfect cubes}=3=\{1,8,27\}
\displaystyle \text{Prime numbers}=15=\{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}
\displaystyle \text{Therefore,}
\displaystyle \text{(i) Probability (divisible by }5)=\frac{10}{50}=\frac{1}{5}
\displaystyle \text{(ii) Probability (a perfect cube)}=\frac{3}{50}
\displaystyle \text{(iii) Probability (a prime number)}=\frac{15}{50}=\frac{3}{10}

\displaystyle \textbf{Question 11: } \text{Cards bearing numbers }2,4,6,8,10,12,14,16,18\text{ and }20\text{ are} \\ \text{kept in a bag. A card is drawn at random from the bag. Find the probability of} \\ \text{getting a card which is}~\text{[ICSE 2018]}
\displaystyle (i)~\text{a prime number}
\displaystyle (ii)~\text{a number divisible by }4
\displaystyle (iii)~\text{a number that is a multiple of }6
\displaystyle (iv)~\text{an odd number.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of outcomes}=10
\displaystyle \text{(i) Prime number}=\{2\}
\displaystyle \text{Number of favourable outcomes}=1
\displaystyle \therefore~\text{Required probability}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{1}{10}
\displaystyle \text{(ii) Number divisible by }4=\{4,8,12,16,20\}
\displaystyle \text{Number of favourable outcomes}=5
\displaystyle \therefore~\text{Required probability}=\frac{5}{10}=\frac{1}{2}
\displaystyle \text{(iii) Number which is a multiple of }6=\{6,12,18\}
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \therefore~\text{Required probability}=\frac{3}{10}
\displaystyle \text{(iv) Number of favourable outcomes (odd numbers)}=0
\displaystyle \text{since, there is no card bearing odd number.}
\displaystyle \therefore~\text{Required probability}=\frac{0}{10}=0

\displaystyle \textbf{Question 12.}~\text{A box consists of 4 red, 5 black and 6 white balls. One ball is} \\ \text{drawn at random. Find the probability that the ball drawn is}~\text{[ICSE 2019]}
\displaystyle (i)~\text{black}\qquad (ii)~\text{red or white}
\displaystyle \text{Answer:}
\displaystyle \text{We have }4\text{ Red, }5\text{ Black and }6\text{ White balls.}
\displaystyle \therefore~\text{Total number of balls}=4+5+6=15
\displaystyle \text{(i) }P(\text{black})=\frac{\text{Number of black balls}}{\text{Total number of balls}}
\displaystyle =\frac{5}{15}=\frac{1}{3}
\displaystyle \text{(ii) }P(\text{red or white})=P(\text{red})+P(\text{white})
\displaystyle =\frac{\text{Number of red balls}}{\text{Total number of balls}}+\frac{\text{Number of white balls}}{\text{Total number of balls}}
\displaystyle =\frac{4}{15}+\frac{6}{15}=\frac{10}{15}=\frac{2}{3}

\displaystyle \textbf{Question 13.}~\text{There are 25 discs numbered 1 to 25. They are put in a closed box} \\ \text{and shaken thoroughly. A disc is drawn at random from the box. Find the} \\ \text{probability that the number on the disc is}~\text{[ICSE 2019]}
\displaystyle (i)~\text{an odd number}
\displaystyle (ii)~\text{divisible by 2 and 3 both}
\displaystyle (iii)~\text{a number less than }16.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Total number of discs in a box, }n(S)=25
\displaystyle \text{Odd number of discs, }E=\{1,3,5,7,9,11,13,15,17,19,21,23,25\}
\displaystyle \therefore~n(E)=13
\displaystyle \therefore~\text{Probability of getting odd number}=\frac{n(E)}{n(S)}=\frac{13}{25}
\displaystyle \text{(ii) Number divisible by }2\text{ and }3\text{ i.e. }6
\displaystyle E_{1}=\{6,12,18,24\}
\displaystyle \therefore~n(E_{1})=4
\displaystyle \therefore~\text{Probability of getting the number divisible by }2\text{ and }3=\frac{n(E_{1})}{n(S)}=\frac{4}{25}
\displaystyle \text{(iii) Number less than }16,
\displaystyle E_{2}=\{1,2,\ldots,15\}
\displaystyle \therefore~n(E_{2})=15
\displaystyle \therefore~\text{Probability of getting a number less than }16=\frac{n(E_{2})}{n(S)}=\frac{15}{25}=\frac{3}{5}

\displaystyle \textbf{Question 14.}~\text{Each of the letters of the word }\text{`AUTHORIZES'}\text{ is written on } \\ \text{identical circular disc and put in a bag. They are well shuffled. If a disc is} \\ \text{drawn at random from the bag, what is the probability that the letter is}~\text{[ICSE 2020]}
\displaystyle (i)~\text{a vowel?}
\displaystyle (ii)~\text{one of the first 9 letters of the English alphabet which appears in the given word?}
\displaystyle (iii)~\text{one of the last 9 letters of the English alphabet which appears in the given word?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, word `AUTHORIZES', so total number of identical circular discs}=10
\displaystyle \text{(i) Here vowels are }A,U,O,I,E
\displaystyle \text{Total number of vowels}=5
\displaystyle \therefore~\text{Required probability}=\frac{5}{10}=\frac{1}{2}
\displaystyle \text{(ii) One of the first }9\text{ letters of the English alphabet which appears in the given word are }A,H,I,E.
\displaystyle \therefore~\text{Required probability}=\frac{4}{10}=\frac{2}{5}
\displaystyle \text{(iii) One of the last }9\text{ letters of the English alphabet which appear in the given word are }U,T,R,Z,S.
\displaystyle \therefore~\text{Required probability}=\frac{5}{10}=\frac{1}{2}

\displaystyle \textbf{Question 15.}~\text{The probability of getting a number divisible by 3 in throwing a dice} \\ \text{is}~\text{[ICSE Semester II 2022]}
\displaystyle (a)~\frac{1}{6}\qquad (b)~\frac{1}{3}\qquad (c)~\frac{1}{2}\qquad (d)~\frac{2}{3}
\displaystyle \text{Answer:}
\displaystyle \text{(b) In a die, }3\text{ and }6\text{ are divisible by }3.
\displaystyle \text{So, probability of getting a number divisible by }3=\frac{2}{6}=\frac{1}{3}.

\displaystyle \textbf{Question 16.}~\text{A bag contains 5 white, 2 red and 3 black balls. A ball is drawn at random.} \\ \text{What is the probability that the ball drawn is a red ball?}~\text{[ICSE Semester II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of balls in the bag}=5+2+3=10
\displaystyle \text{Number of red balls}=2
\displaystyle \therefore~\text{Probability of getting a red ball}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{2}{10}=\frac{1}{5}.

\displaystyle \textbf{Question 17.}~\text{A letter of the word }\text{`SECONDARY'}\text{ is selected at random.} \\ \text{What is the probability that the letter selected is not a vowel?}~\text{[ICSE Semester II 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of letters in the word `SECONDARY'}=9
\displaystyle \text{Total number of consonants present in the given word}=6
\displaystyle \text{So, probability that the letter selected is not a vowel in the given word}=\frac{6}{9}=\frac{2}{3}.

\displaystyle \textbf{Question 18.}~\text{If the probability of a player winning a game is }0.56,\text{ then the} \\ \text{probability of his losing this game is}~\text{[ICSE Semester 2022]}
\displaystyle (a)~0.56\qquad (b)~1\qquad (c)~0.44\qquad (d)~0
\displaystyle \text{Answer:}
\displaystyle \text{(c) Here, }P(\text{a player winning a game})=0.56
\displaystyle \therefore~P(\text{losing this game})=1-0.56=0.44.

\displaystyle \textbf{Question 19.}~\text{A bag contains 25 cards, numbered through 1 to 25. A card is drawn} \\ \text{at random. What is the probability that the number on the card drawn is}~\text{[ICSE 2023]}
\displaystyle (i)~\text{multiple of 5}\qquad (ii)~\text{a perfect square}\qquad (iii)~\text{a prime number?}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Cards which are multiples of }5\text{ are }5,10,15,20,25.
\displaystyle \therefore~\text{Probability of drawn card to be a multiple of }5\text{ is}
\displaystyle P_{1}=\frac{\text{Number of card multiples of }5}{\text{Total number of cards}}
\displaystyle P_{1}=\frac{5}{25}\Rightarrow P_{1}=\frac{1}{5}
\displaystyle \text{(ii) Perfect squares in }1\text{ to }25\text{ are }1,4,9,16,25.
\displaystyle \text{Let probability of drawn card to be a perfect square is }P_{2}.
\displaystyle P_{2}=\frac{\text{Number of perfect squares}}{\text{Total numbers of cards}}
\displaystyle \Rightarrow P_{2}=\frac{5}{25}=\frac{1}{5}
\displaystyle \text{(iii) Prime numbers in }1\text{ to }25\text{ are }2,3,5,7,11,13,17,19,23.
\displaystyle \text{Let probability of drawn card to be a prime number is }P_{3}.
\displaystyle \Rightarrow P_{3}=\frac{\text{Number of primes}}{\text{Total numbers of cards}}
\displaystyle \Rightarrow P_{3}=\frac{9}{25}

\displaystyle \textbf{Question 20.}~\text{A box contains some green, yellow and white tennis balls. The probability} \\ \text{of selecting a green ball is }\frac{1}{4}\text{ and that of selecting a yellow ball is }\frac{1}{3}.\text{ If the box contains} \\ \text{10 white balls, then find}~\text{[ICSE 2023]}
\displaystyle (i)~\text{total number of balls in the box}\qquad (ii)~\text{probability of selecting a white ball.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the total number of balls in the box be }x.
\displaystyle \therefore~n(S)=x
\displaystyle \text{Let }A\text{ be the event of green ball}\Rightarrow P(A)=\frac{1}{4}
\displaystyle \text{and }B\text{ be the event of yellow ball}\Rightarrow P(B)=\frac{1}{3}
\displaystyle \text{and let }C\text{ be the event of white ball.}
\displaystyle \text{We know that sum of all probabilities of all events in a sample space is equal to }1
\displaystyle \therefore~P(A)+P(B)+P(C)=1
\displaystyle \Rightarrow \frac{1}{4}+P(C)+\frac{1}{3}=1
\displaystyle \therefore~P(C)=1-\frac{1}{4}-\frac{1}{3}=1-\frac{7}{12}=\frac{5}{12}\quad \ldots (i)
\displaystyle \therefore~n(S)=x\text{ and }n(C)=10\quad [\text{given}]
\displaystyle \therefore~P(C)=\frac{10}{x}\quad \ldots (ii)
\displaystyle \text{From Eqs. (i) and (ii), }\frac{10}{x}=\frac{5}{12}\Rightarrow x=24
\displaystyle \text{(i) Hence, the total number of balls in the box is }24.
\displaystyle \text{(ii) Probability of selecting a white ball is }\frac{5}{12}.

\displaystyle \textbf{Question 21.}~\text{The probability of the Sun rising from the East is }P(S).\text{ The} \\ \text{value of }P(S)\text{ is}~\text{[ICSE  2023]}
\displaystyle (a)~P(S)=0\qquad (b)~P(S)<0\qquad (c)~P(S)=1\qquad (d)~P(S)>1
\displaystyle \text{Answer:}
\displaystyle \text{(c) Since, Sun rising from the East is a sure event and}
\displaystyle \text{probability of a sure event is }1.
\displaystyle \therefore~\text{The value of }P(S)=1

\displaystyle \textbf{Question 22.}~\text{A letter is chosen at random from all the letters of the English alphabets.} \\ \text{The probability that the letter chosen is a vowel, is}~\text{[ICSE 2023]}
\displaystyle (a)~\frac{4}{26}\qquad (b)~\frac{5}{26}\qquad (c)~\frac{21}{26}\qquad (d)~\frac{5}{24}
\displaystyle \text{Answer:}
\displaystyle \text{(b) We know,}
\displaystyle \text{the number of English alphabets}=26
\displaystyle \text{the number of vowels in English alphabets}=5
\displaystyle \therefore~\text{The probability that the letter chosen is a vowel}
\displaystyle =\frac{\text{The number of vowels in English alphabets}}{\text{The number of English alphabets}}=\frac{5}{26}

\displaystyle \textbf{Question 23. } \text{Assertion (A)}~\text{A die is thrown once and the probability of getting an even} \\ \text{number is }\frac{2}{3}.
\displaystyle \text{Reason (R)}~\text{The sample space for even numbers on a die is }\{2,4,6\}.~\text{[ICSE 2024]}
\displaystyle (a)~\text{A is true but R is false.}
\displaystyle (b)~\text{A is false but R is true.}
\displaystyle (c)~\text{Both A and R are true.}
\displaystyle (d)~\text{Both A and R are false.}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Assertion (A)}~\text{When a die is thrown once, then the}
\displaystyle \text{possible outcomes are }6
\displaystyle \text{i.e.}\quad S=\{1,2,3,4,5,6\}
\displaystyle \text{Even number}=\{2,4,6\}
\displaystyle \therefore~\text{Favourable outcomes}=3
\displaystyle \text{Thus, the probability of getting an even number}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}
\displaystyle =\frac{3}{6}=\frac{1}{2}
\displaystyle \text{So, Assertion is false but Reason is true.}

\displaystyle \textbf{Question 24.}~\text{A box contains tokens numbered 5 to 16. A token is drawn at random.} \\ \text{Find the probability that the token drawn bears a number divisible by} ~\text{[ICSE 2024]}
\displaystyle (i)~5\qquad (ii)~\text{Neither by 2 nor by 3}
\displaystyle \text{Answer:}
\displaystyle \text{Given, the box contains tokens numbered from }5\text{ to }16.
\displaystyle \text{Total number of possible outcomes}=12
\displaystyle \text{(i) Let }E\text{ be the event of getting a number divisible by }5.
\displaystyle \text{Then, the outcomes favourable to }E\text{ are }5,10,15.
\displaystyle \therefore~\text{Number of outcomes favourable to }E=3.
\displaystyle \therefore~P(\text{token number is divisible by }5)=\frac{3}{12}=\frac{1}{4}.
\displaystyle \text{(ii) Let }F\text{ be the event of getting a number neither divisible by }2\text{ nor by }3.
\displaystyle \text{Then, the outcomes favourable to }F\text{ are }5,7,11,13.
\displaystyle \therefore~\text{Number of outcomes favourable to }F=4
\displaystyle \therefore~P(F)=\frac{4}{12}=\frac{1}{3}

\displaystyle \textbf{Question 25.}~\text{The following letters }A,D,M,N,O,S,U,Y\text{ of the English alphabet} \\ \text{are written on separate cards and put in a box. The cards are well shuffled and one} \\ \text{card is drawn at random. What is the probability that the card drawn is a letter of} \\ \text{the word}~\text{[ICSE 2024]}
\displaystyle (i)~\text{MONDAY?}
\displaystyle (ii)~\text{which does not appear in MONDAY?}
\displaystyle (iii)~\text{which appears both in SUNDAY and MONDAY?}
\displaystyle \text{Answer:}
\displaystyle \text{Given, letters are }A,D,M,N,O,S,U,Y
\displaystyle \text{Total number of letters}=8
\displaystyle \text{(i) The number of letters in word `MONDAY'}=6
\displaystyle \text{Number of favourable outcomes}=6
\displaystyle \therefore~\text{Probability that the card drawn is a letter of the word `MONDAY'}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}
\displaystyle =\frac{6}{8}=\frac{3}{4}
\displaystyle \text{(ii) The letters which do not appear in MONDAY are }S\text{ and }U.
\displaystyle \therefore~\text{Probability that the card drawn is a letter which does not appear in `MONDAY'}=\frac{2}{8}=\frac{1}{4}
\displaystyle \text{(iii) Number of letters which appear in MONDAY and SUNDAY}=4
\displaystyle \therefore~\text{Probability that the card drawn is a letter which appears in both SUNDAY and MONDAY}
\displaystyle =\frac{4}{8}=\frac{1}{2}

\displaystyle \textbf{Question 26.}~\text{Two coins are tossed once. Find the probability of getting}~\text{[ICSE 2012]}
\displaystyle (i)~2\text{ heads.}\qquad (ii)~\text{at least }1\text{ tail.}
\displaystyle \text{Answer:}
\displaystyle \text{When two coins are tossed once. Then, possible outcomes}=\{HH,HT,TH,TT\}
\displaystyle \therefore~\text{Total number of outcomes}=4
\displaystyle \text{(i) Favourable outcomes}= \text{getting }2\text{ heads}=\{HH\}
\displaystyle \text{Number of favourable outcomes}=1
\displaystyle \therefore~\text{Probability of getting }2\text{ heads}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{1}{4}
\displaystyle \text{(ii) Favourable outcomes}= \text{getting atleast }1\text{ tail}=\{HT,TH,TT\}
\displaystyle \text{Number of favourable outcomes}=3
\displaystyle \therefore~\text{Probability of getting atleast }1\text{ tail}
\displaystyle =\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}=\frac{3}{4}

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