Exercise 22
Question 1
A bag contains a red ball, a blue ball and a yellow ball, all the balls
being of the same size. Anjali takes out a ball from the bag without looking
into it. What is the probability that she takes out
(i) yellow ball ?
(ii) red ball ?
(iii) blue ball ?
Sol :
Number of balls in the bag = 3.
(i) Probability of yellow ball $=\frac{1}{3}$
(ii) Probability of red ball $=\frac{1}{3}$
(iii) Probability of blue ball $=\frac{1}{3}$
Question 2
A box contains 600 screws, one-tenth are rusted. One screw is taken out at
random from this box. Find the probability that it is a good screw.
Sol :
Number of total screws = 600
Rusted screws$=\frac{1}{10}$ of $600=60$
∴ Good screws = 600 – 60 = 540
Probability of a good screw
$P(E)=\frac{\text { Number of farourable outcome }}{\text { Number of possible
outcome }}$
$=\frac{540}{800}$
$=\frac{9}{10}$
Question 3
In a lottery, there are 5 prized tickets and 995 blank tickets. A person
buys a lottery ticket. Find the probability of his winning a prize.
Sol :
Number of prized tickets = 5
Number of blank tickets = 995
Total number of tickets = 5 + 995 = 1000
Probability of prized ticket
$P(E)=\frac{N \text { umber of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{5}{1000}$
$=\frac{1}{200}$
Question 4
12 defective pens are accidentally mixed with 132 good ones. It is not
possible to just look at a pen and tell whether or not it is defective. One
pen is taken out at random from this lot. Determine the probability that the
pen taken out is a good one.
Sol :
Number of defective pens = 12
Number of good pens = 132
Total number of pens =12 + 132 = 144
Probability of good pen
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{132}{144}$
$=\frac{11}{12}$
Question 5
If the probability of winning a game is $\frac{5}{11}$ , what is the
probability of losing ?
Sol :
Probability of winning game $=\frac{5}{11}$
$\Rightarrow P(E)=\frac{5}{11}$
We know that $P(E)+P(\bar{E})=1$
where P (E) is the probability of losing the game.
$\frac{5}{11}+P(\bar{E})=1$
$\Rightarrow P(\bar{E})=1-\frac{5}{11}$
$=\frac{11-5}{11}$
$=\frac{6}{11}$
Question 6
Two players, Sania and Sonali play a tennis match. It is known that the
probability of Sania winning the match is 0.69. What is the probability of
Sonali winning ?
Sol :
Probability of Sania’s winning the game = 0.69
Let P (E) be the probability of Sania’s winning the game
and $P(\bar{E})$ be the probability of Sania’s losing
the game or probability of Sonali, winning the game
$P(E)+P(\bar{E})=1$
$\Rightarrow 0.69+P(\bar{E})=1$
$\Rightarrow P(\bar{E})=1-0.69=0.31$
Hence probability of Sonali’s winning the game = 0.31
Question 7
A bag contains 3 red balls and 5 black balls. A ball is drawn at random’
from in bag. What is the probability that the ball drawn is .
(i) red ?
(ii) not red ?
Sol :
Number of red balls = 3
Number of black balls = 5
Total balls = 3 + 5 = 8
Let P (E) be the probability of red balls,
then $P(\bar{E})$ will be the probability of not red balls.
$P(E)+P(\bar{E})=1$
(i) But $P(E)=\frac{N_{\text {umber }} \text { of favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{3}{8}$
(ii) $\mathrm{P}(\bar{E})=1-\mathrm{P}(\mathrm{E})$
$=1-\frac{6}{11}$
$=\frac{8-3}{8}$
$=\frac{5}{8}$
Question 8
There are 40 students in Class X of a school of which 25 are girls and
the.others are boys. The class teacher has to select one student as a class
representative. She writes the name of each student on a separate card, the
cards being identical. Then she puts cards in a bag and stirs them
thoroughly. She then draws one card from the bag. What is the probability
that the name written on the card is the name of
(i) a girl ?
(ii) a boy ?
Sol :
Number of total students = 40
Number of girls = 25
Number of boys = 40 – 25 = 15
(i) Probability of a girl
$P(E)=\frac{\text { Number } \text { of favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{25}{40}$
$=\frac{5}{8}$
(ii) Probability of a boy
$P(E)=\frac{\text { Number } \text { of favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{15}{40}$
$=\frac{3}{8}$
Question 9
A letter is chosen from the word ‘TRIANGLE’. What is the probability that
it is a vowel ?
Sol :
There are three vowels: I, A, E
∴ The number of letters in the word ‘TRIANGLE’ = 8.
Probability of vowel
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{3}{8}$
Question 10
A letter of English alphabet is chosen at random. Determine the probability
that the letter is a consonant.
Sol :
No. of English alphabet = 26
No. of vowel = 5
No. of constant = 25 – 5 = 21
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of possible
outcome }}$
$=\frac{21}{26}$
Question 11
A bag contains 5 black, 7 red and 3 white balls. A ball is drawn at random
from the bag, find the probability that the ball drawn is:
(i) red
(ii) black or white
(iii) not black.
Sol :
In a bag,
Number of black balls = 5
Number of red balls = 7
and number of white balls = 3
Total number of balls in the bag
= 5 + 7 + 3 = 15
(i) Probability of red balls
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of possible
outcome }}$
$=\frac{7}{15}$
(ii) Probability of black or white balls
$P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of possible
outcome }}$
$=\frac{5+3}{15}$
$=\frac{8}{15}$
(iii) Probability of not black balls
$P(E)=\frac{\text { Number of favourable outcome }}{\text { Number of
possible outcome }}$
$=\frac{7+3}{15}$
$=\frac{10}{15}$
$=\frac{2}{3}$
Question 12
A box contains 7 blue, 8 white and 5 black marbles. If a marble is drawn at
random from the box, what is the probability that it will be
(i) black?
(ii) blue or black?
(iii) not black?
(iv) green?
Sol :
Total number of marbles in the box
= 7 + 8 + 5 = 20
Since, a marble is drawn at random from the box
(i) Probability (of a black Marble)
$P(E)=\frac{N{\text {umber }} \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{5}{20}$
$=\frac{1}{4}$
(ii) Probability (of a blue or black marble)
$P(E)=\frac{\text { Number of favourable ontcome }}{\text { Number of possible
outcome }}$
$=\frac{7+5}{20}$
$=\frac{12}{20}$
$=\frac{3}{5}$
(iii) Probability (of not black marble)
= 1 – P (of black 1)
$=1-\frac{1}{4}$
$=\frac{4-1}{4}$
$=\frac{3}{4}$
(iv) P (of a green marble) = 0
(∴ Since, a box does not contain a green marble,
so the probability of green marble will be zero)
Question 13
A bag contains 6 red balls, 8 white balls, 5 green balls and 3 black balls.
One ball is drawn at random from the bag. Find the probability that the ball
is :
(i) white
(ii) red or black
(iii) not green
(iv) neither white nor black.
Sol :
In a bag,
Number of red balls = 6
Number of white balls = 8
Number of green balls = 5
and number of black balls = 3
Total number of balls in the bag
= 6 + 8 + 5 + 3 = 22
(i) Probability of white balls
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{8}{22}$
$=\frac{4}{11}$
(ii) Probability of red or black balls
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{6+3}{22}$
$=\frac{9}{22}$
(iii) Probability of not green balls i.e. having red, white and black balls.
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{6+8+3}{22}$
$=\frac{17}{22}$
(iv) Probability of neither white nor black balls i.e. red and green balls
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{6+5}{22}$
$=\frac{11}{22}$
$=\frac{1}{2}$
Question 14
A piggy bank contains hundred 50 p coins, fifty Rs 1 coins, twenty Rs 2
coins and ten Rs 5 coins. It is equally likely that one of the coins will
fall down when the bank is turned upside down, what is the probability that
the coin
(i) will be a 50 p coin?
(ii) will not be Rs 5 coin?
Sol :
In a piggy bank, there are
100, 50 p coin
50, Rs 1 coin
20, Rs 2 coin
10, Rs 5 coin
Total coins = 100 + 50 + 20 + 10 = 180
One coin is drawn at random Probability of
(i) 50 p coins $=\frac{100}{180}$
$=\frac{5}{9}$
(ii) Will not be Rs 5 coins
= 100 + 50 + 20 = 170
Probability$=\frac{170}{180}=\frac{17}{18}$
Question 15
A carton consists of 100 shirts of which 88 are good, 8 have minor defects
and 4 have major defects. Peter, a trader, will only accept the shirts which
are good, but Salim, another trader, will only reject the shirts which have
major defects. One shirts is drawn at random from the carton. What is the
probability that
(i) it is acceptable to Peter ?
(ii) it is acceptable to Salim ?
Sol :
In a carton, there the 100 shirts.
Among these number of shirts which are good = 88
number of shirts which have minor defect = 8
number of shirt which have major defect = 4
Total number of shirts = 88 + 8 + 4 = 100
Peter accepts only good shirts i.e. 88
Salim rejects only shirts which have major defect i.e. 4
(i) Probability of good shirts which are acceptable to Peter
$P(E)=\frac{\text { Number } \text { of favourable outocme}}{\text { Number of
possible outcome }}$
$=\frac{88}{100}$
$=\frac{22}{25}$
(ii) Probability of shirts acceptable to Salim
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of possible
outcome }}$
$=\frac{88+8}{100}$
$=\frac{96}{100}$
$=\frac{24}{25}$
Question 16
A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?
Sol :
Dice is thrown once
Sample space = {1, 2, 3, 4, 5, 6}
(i) No. of ways in favour = 3
(∵ Even numbers are 2, 4, 6)
Total ways = 6
Probability $=\frac{3}{6}=\frac{1}{2}$
(ii) No. of ways in favour = 4
(Numbers greater than 2 are 3, 4, 5, 6)
Total ways = 6
Probability $=\frac{4}{6}=\frac{2}{2}$
Question 17
In a single throw of a die, find the probability of getting:
(i) an odd number
(ii) a number less than 5
(iii) a number greater than 5
(iv) a prime number
(v) a number less than 8
(vi) a number divisible by 3
(vii) a number between 3 and 6
(viii) a number divisible by 2 or 3.
Sol :
A die is thrown and on its faces, numbers 1 to 6 are written.
Total numbers of possible outcomes = 6
(i) Probability of an odd number,
odd number are 1, 3 and 5
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of possible
outcome }}$
$=\frac{3}{6}$
$=\frac{1}{2}$
(ii) A number less them 5 are 1, 2, 3, 4
Probability of a number less than 5 is
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of
possible outcome }}$
$=\frac{4}{6}$
$=\frac{2}{3}$
(iii) A number greater than 5 is 6
Probability of a number greater than 5 is
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of
possible outcome }}$
$=\frac{1}{6}$
(iv) Prime number is 2, 3, 5
Probability of a prime number is
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of
possible outcome }}$
$=\frac{3}{6}$
$=\frac{1}{2}$
(v) Number less than 8 is nil
P (E) = 0
(vi) A number divisible by 3 is 3, 6
Probability of a number divisible by 3 is
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of
possible outcome }}$
$=\frac{2}{6}$
$=\frac{1}{3}$
(vii) Numbers between 3 and 6 is 4, 5
Probability of a number between 3 and 6 is
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of
possible outcome }}$
$=\frac{2}{6}$
$=\frac{1}{3}$
(viii) Numbers divisible by 2 or 3 are 2, 4 or 3,
Probability of a number between 2 or 3 is
$=\frac{2}{6}$
$=\frac{1}{3}$
Question 18
A die has 6 faces marked by the given numbers as shown below:
The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than – 3.
(iii) the smallest integer ?
Sol :
Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourables n(E) = 3
Probability $=\frac{n(E)}{n(S)}$
$=\frac{3}{6}$
$=\frac{1}{2}$
(ii) Integer greater than -3
= (1, 2, 3, -1, -2)
No. of favourables n(E) = 5
Probability =$\frac{n(E)}{n(S)}$
$=\frac{5}{6}$
(iii) Smallest integer = -3
No. of favourables n(E) = 1
Probability $=\frac{n(E)}{n(S)}$
$=\frac{1}{6}$
A game of chance consists of spinning an arrow which comes to rest
pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (shown in the
adjoining figure) and these are equally likely outcomes. What is the
probability that it will point at
(i) 8 ?
(ii) an odd number ?
(iii) a number greater than 2?
(iv) a number less than 9?
Sol :
On the face of a game, numbers 1 to 8 is shown.
Possible outcomes = 8
(i) Probability of number 8 will be
$P(E)=\frac{\text { Number of favourable ootcome }}{\text { Number of possible
outcome }}$
$=\frac{1}{8}$
(ii) Odd number are 1, 3, 5, 7
Probability of a number which is an odd will be
$P(E)=\frac{\text { Number of favourable ootcome }}{\text { Number of
possible outcome }}$
$=\frac{4}{8}$
$=\frac{1}{2}$
(iii) A number greater than 2 are 3, 4, 5, 6, 7, 8 which are 6
Probability of number greater than 2 will be
$P(E)=\frac{\text { Number of favourable ootcome }}{\text { Number of
possible outcome }}$
$=\frac{6}{8}$
$=\frac{3}{4}$
(iv) A number less than 9 is 8.
Probability of a number less than 9 will be
$P(E)=\frac{\text { Number of favourable ootcome }}{\text { Number of
possible outcome }}$
$=\frac{8}{8}$
Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.
Sol :
In January, there are 31 days and in an ordinary year,
there are 365 days but in a leap year, there are 366 days.
(i) In January of an ordinary year, there are 31 days i.e. 4 weeks and 3
days.
Probability of Monday will be$=\frac{3}{7}$
(ii) In January of a leap year, there are 31 days i.e. 4 weeks and 3 days
Probability of Monday will be $=\frac{3}{7}$
Find the probability that the month of February may have 5 Wednesdays
in
(i) a leap year
(ii) a non-leap year.
Sol :
In the month of February, there are 29 days in a leap year
while 28 days in a non-leap year,
(i) In a leap year, there are 4 complete weeks and 1 day
Probability of Wednesday = P (E) $=\frac{1}{7}$
(ii) and in a non leap year, there are 4 complete weeks and 0 days
Probability of Wednesday P (E) $=\frac{0}{7}=0$
Sixteen cards are labelled as a, b, c,…, m, n, o, p. They are put in a
box and shuffled. A boy is asked to draw a card from the box. What is the
probability that the card drawn is:
(i) a vowel
(ii) a consonant
(iii) none of the letters of the word median.
Sol :
Here, sample space (S) = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p)
∴n(S) = 16
(i) Vowels (V) = {a, e, i, o}
∴n(V) = 4
∴P(a vowel) $=\frac{n(V)}{n(S)}=\frac{4}{16}=\frac{1}{4}$
(ii) Consonants (C) = {b, c, d, f, g, h, j, k, l, m, n, p}
∴n(C) = 12
∴P (a consonant)$=\frac{n(C)}{n(S)}=\frac{12}{16}=\frac{3}{4}$
(iii) None of the letters of the word MEDIAN (N) = {b, c, f, g, h, j, k, l,
o, p)
∴n(N) = 10
∴P (N) $=\frac{n(N)}{n(S)}=\frac{10}{16}=\frac{5}{8}$
An integer is chosen between 0 and 100. What is the probability that it
is
(i) divisible by 7?
(ii) not divisible by 7?
Sol :
Integers between 0 and 100 = 99
(i) Number divisible by 7 are
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98 = 14
Probability$=\frac{14}{99}$
(ii) Not divisible by 7 are 99 – 14 = 85
Probability $=\frac{85}{99}$
Cards marked with numbers 1, 2, 3, 4, 20 are well shuffled and a card is
drawn at random.
What is the probability that the number on the card is
(i) a prime number
(ii) divisible by 3
(iii) a perfect square ? (2010)
Sol :
Number cards is drawn from 1 to 20 = 20
One card is drawn at random
No. of total (possible) events = 20
(i) The card has a prime number
The prime number from 1 to 20 are 2, 3, 5, 7, 11, 13, 17, 19
Actual No. of events = 8
$P(E)=\frac{\text { Number of actual events }}{\text { Number of total events
}}$
$=\frac{8}{20}$
$=\frac{2}{5}$
(ii) Numbers divisible by 3 are 3, 6, 9, 12, 15, 18
No. of actual events = 6
$P(E)=\frac{\text { Number of actual events }}{\text { Number of total
events }}$
$=\frac{6}{20}$
$=\frac{3}{10}$
(iii) Numbers which are perfect squares = 1, 4, 9, 16 = 4
$P(E)=\frac{\text { Number of actual events }}{\text { Number of total
events }}$
$=\frac{4}{20}$
$=\frac{1}{5}$
A box contains 25 cards numbered 1 to 25. A card is drawn from the box at
random. Find the probability that the number on the card is :
(i) even
(ii) prime
(iii) multiple of 6
Sol :
Number of card in a box = 25 numbered 1 to 25
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24
i.e. number of favourable outcomes = 12
Probability of an even number will be
$P(E)=\frac{12}{25}$
(ii) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23
i.e. number of primes = 9
Probability of primes will be
$P(E)=\frac{ \text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{9}{25}$
(iii) Multiples of 6 are 6, 12, 18, 24
Number of multiples = 4
Probability of multiples of 6 will be
$P(E)=\frac{ \text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{4}{25}$
A box contains 15 cards numbered 1, 2, 3,…..15 which are mixed
thoroughly. A card is drawn from the box at random. Find the probability
that the number on the card is :
(i) Odd
(ii) prime
(iii) divisible by 3
(iv) divisible by 3 and 2 both
(v) divisible by 3 or 2
(vi) a perfect square number.
Sol :
Number of cards in a box =15 numbered 1 to 15
(i) Odd numbers are 1, 3, 5, 7, 9, 11, 13, 15
Number of odd numbers = 8
Probability of odd numbers will be
$P(E)=\frac{\text { Number of favourable outcome}}{\text { Number of possible
outcome }}$
$=\frac{8}{15}$
(ii) Prime number are 2, 3, 5, 7, 11, 13
Number of primes is 6
Probability of prime number will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{6}{15}$
$=\frac{2}{5}$
(iii) Numbers divisible by 3 are 3, 6, 9, 12, 15
which are 5 in numbers
Probability of number divisible by 3 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{5}{15}$
$=\frac{1}{3}$
(iv) Divisible by 3 and 2 both are 6, 12
which are 2 in numbers.
Probability of number divisible by 3 and 2
Both will be$=\frac{2}{15}$
(v) Numbers divisible by 3 or 2 are
2, 3, 4, 6, 8, 9, 10, 12, 14, 15 which are 10 in numbers
Probability of number divisible by 3 or 2 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{3}{15}$
$=\frac{1}{5}$
A box contains 19 balls bearing numbers 1, 2, 3,…., 19. A ball is drawn
at random from the box. Find the probability that the number on the ball
is :
(i) a prime number
(ii) divisible by 3 or 5
(iii) neither divisible by 5 nor by 10
(iv) an even number.
Sol :
In a box, number of balls = 19 with number 1 to 19.
A ball is drawn
Number of possible outcomes = 19
(i) Prime number = 2, 3, 5, 7, 11, 13, 17, 19
which are 8 in number
Probability of prime number will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{8}{19}$
(ii) Divisible by 3 or 5 are 3, 5, 6, 9, 10, 12, 15, 18
which are 8 in number
Probability of number divisible by 3 or 5 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{8}{19}$
(iii) Numbers which are neither divisible by 5 nor by 10 are
1, 2, 3, 4, 6, 7, 8, 9, 11, 12,
13, 14, 16, 17, 18, 19
which are 16 in numbers
Probability of there number will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{16}{19}$
(iv) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18
which are 9 in numbers.
Probability of there number will be
Number of favourable outcome
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{9}{19}$
Cards marked with numbers 13, 14, 15, …, 60 are placed in a box and mixed
thoroughly. One card is drawn at random from the box. Find the probability
that the number on the card drawn is
(i) divisible by 5
(ii) a perfect square number.
Sol :
Number of card bearing numbers 13,14,15, … 60 = 48
One card is drawn at random.
(i) Card divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability $=\frac{10}{48}$
$=\frac{5}{24}$
(ii) A perfect square = 16, 25, 36, 49 = 4
Probability$=\frac{4}{48}$
$=\frac{1}{12}$
Tickets numbered 3, 5, 7, 9,…., 29 are placed in a box and mixed
thoroughly. One ticket is drawn at random from the box. Find the
probability that the number on the ticket is
(i) a prime number
(ii) a number less than 16
(iii) a number divisible by 3.
Sol :
In a box there are 14 tickets with number
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29
Number of possible outcomes = 14
(i) Prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29
which are 9 in number
Probability of prime will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{9}{14}$
(ii) Number less than 16 are 3, 5, 7, 9, 11, 13, 15
which are 7 in numbers,
Probability of number less than 16 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{7}{14}$
$=\frac{1}{2}$
(iii) Numbers divisible by 3 are 3, 9, 15, 21, 27
which are 5 in number
Probability of number divisible by 3 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{5}{14}$
A box contains 90 discs which are numbered from 1 to 90. If one disc is
drawn at random from the box, find the probability that it bears
(i) a two-digit number
(ii) a perfect square number
(iii) a number divisible by 5.
Sol :
There are 90 discs in a box containing numbered from 1 to 90.
Number of possible outcomes = 90
(i) Two digit numbers are 10 to 90 which are 81 in numbers.
Probability of two digit number will be
$=\frac{81}{90}$
$=\frac{9}{10}$
(ii) Perfect squares are 1, 4, 9, 16, 25, 36,49, 64, 81
which are 9 in numbers.
Probability of square will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{9}{90}$
$=\frac{1}{10}$
(iii) Number divisible by 5 are
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
which are 18 in numbers.
Probability of number divisible by 5 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{18}{90}$
$=\frac{1}{5}$
Cards marked with numbers 2 to 101 are placed in a box and mixed
thoroughly. One card is drawn at random from this box. Find the
probability that the number on the card is
(i) an even number
(ii) a number less than 14
(iii) a number which is a perfect square
(iv) a prime number less than 30.
Sol :
Number of cards with numbered from 2 to 101 are placed in a box
Number of possible outcomes = 100 one card is drawn
(i) Even numbers are 2, 4, 6, 8, 10, 12, 14, 16,….., 96, 98, 100
which are 50 in numbers.
Probability of even number will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{50}{100}$
$=\frac{1}{2}$
(ii) Numbers less than 14 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
which are 12 in numbers
Probability of number less than 14 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{12}{100}$
$=\frac{3}{25}$
(iii) Perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100 which are 9 in
numbers
Probability of perfect square number will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{9}{100}$
(iv) Prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 which
are 10 in numbers Probability of prime numbers, less than 30 will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{10}{100}$
$=\frac{1}{10}$
A bag contains 15 balls of which some are white and others are red. If
the probability of drawing a red ball is twice that of a white ball, find
the number of white balls in the bag.
Sol :
In a bag, there are 15 balls.
Some are white and others are red.
Probability of red ball = 2 probability of white ball
Let number of white balls = x
Then, number of red balls = 15 – x
$2 \times \frac{15-x}{15}=\frac{x}{15}$
$\Rightarrow 2(15-x)=x$
$\Rightarrow 30-2 x=x$
$\Rightarrow 30=x+2 x$
$\Rightarrow x=\frac{30}{3}=10$
Number of red balls = 10
and Number of white balls = 15 – 10 = 5
Question 33
A bag contains 6 red balls and some blue balls. If the probability of
drawing a blue ball is twice that of a red ball, find the number of balls
in the bag.
Sol :
In a bag, there are 6 red balls, and some blue balls
Probability of blue ball = 2 × probability of red ball
Let number of blue balls = x
and number of red balls = 6
Total balls = x + 6
Probability of a blue ball = 2
$\Rightarrow \frac{x}{x+6}=2 \times \frac{6}{x+6}$
$\Rightarrow \frac{x}{x+6}=\frac{12}{x+6}$
Number of balls = x + 6 = 12 + 6 = 18
Question 34
A bag contains 24 balls of which x are red, 2x are white and 3x are blue.
A blue is selected at random. Find the probability that it is
(i) white
(ii) not red.
Sol :
In a bag, there are 24 balls
Since, there are x balls red, 2 × balls white and 3 × balls blue
x + 2x + 3x = 24
$\Rightarrow 6 x=24 \Rightarrow x=\frac{24}{6}=4$
Hence, there shall be 4 red balls
$(2 \times 4)=8$ white balls
and $(3 \times 4)=12$ blue balls
(i) $\mathrm{P}$ (white) $=\frac{8}{24}=\frac{1}{3}$
(ii) $\mathrm{P}($ not red $)=\frac{8+12}{24}=\frac{20}{24}=\frac{5}{6}$
$(\because P$ (not red) means probability of either white or blue)
A card is drawn from a well-shuffled pack of 52 cards. Find the
probability of getting:
(i) ‘2’ of spades
(ii) a jack .
(iii) a king of red colour
(iv) a card of diamond
(v) a king or a queen
(vi) a non-face card
(vii) a black face card
(viii) a black card
(ix) a non-ace
(x) non-face card of black colour
(xi) neither a spade nor a jack
(xii) neither a heart nor a red king
Sol :
In a playing card, there are 52 cards
Number of possible outcome = 52
(i) Probability of‘2’ of spade will be
$P(E)=\frac{\text { Number of } \text { favourable } \text { outcome }}{\text
{ Number of possible outcome }}$
$=\frac{1}{52}$
(ii) There are 4 jack card Probability of jack will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text { outcome
}}{\text { Number of possible outcome }}$
$=\frac{4}{52}$
$=\frac{1}{13}$
(iii) King of red colour are 2 in number
Probability of red colour king will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{2}{52}$
$=\frac{1}{26}$
(iv) Cards of diamonds are 13 in number
Probability of diamonds card will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{13}{52}$
$=\frac{1}{4}$
(v) Number of kings and queens = 4 + 4 = 8
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{8}{52}$
$=\frac{2}{13}$
(vi) Non-face cards are = 52 – 3 × 4 = 52 – 12 = 40
Probability of non-face card will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{40}{52}$
$=\frac{10}{13}$
(vii) Black face cards are = 2 × 3 = 6
Probability of black face card will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{6}{52}$
$=\frac{3}{26}$
(viii) No. of black cards = 13 x 2 = 26
Probability of black card will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{26}{52}$
$=\frac{1}{2}$
(ix) Non-ace cards are 12 × 4 = 48
Probability of non-ace card will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{48}{52}$
$=\frac{12}{13}$
(x) Non-face card of black colours are 10 × 2 = 20
Probability of non-face card of black colour will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{20}{52}$
$=\frac{5}{13}$
(xi) Number of card which are neither a spade nor a jack
= 13 × 3 – 3 = 39 – 3 = 36
Probability of card which is neither a spade nor a jack will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{36}{52}$
$=\frac{9}{13}$
(xii) Number of cards which are neither a heart nor a red king
= 3 × 13 = 39 – 1 = 38
Probability of card which is neither a heart nor a red king will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{38}{52}$
$=\frac{19}{26}$
All the three face cards of spades are removed from a well-shuffled pack
of 52 cards. A card is then drawn at random from the remaining pack. Find
the probability of getting
(i) a black face card
(ii) a queen
(iii) a black card
(iv) a heart
(v) a spade
(vi) ‘9’ of black colour
Sol :
In a pack of 52 cards
All the three face cards of spade are = 3
Number of remaining cards = 52 – 3 = 49
One card is drawn at random
(i) Probability of a black face card which are = 6 – 3 = 3
Probability$=\frac{3}{49}$
(ii) Probability of being a queen which are 4 – 1 = 3
Probability $=\frac{3}{49}$
(iii) Probability of being a black card = (26 – 3 = 23)
Probability $=\frac{23}{49}$
(iv) Probability of being a heart $=\frac{13}{49}$
(v) Probability of being a spade = (13 – 3 = 10)
Probability $=\frac{10}{49}$
(vi) Probability of being 9 of black colour (which are 2) $=\frac{2}{49}$
From a pack of 52 cards, a blackjack, a red queen and two black kings
fell down. A card was then drawn from the remaining pack at random. Find
the probability that the card drawn is
(i) a black card
(ii) a king
(iii) a red queen.
Sol :
In a pack of 52 cards, a blackjack, a red queen, two black being felt down.
Then number of total out comes = 52 – (1 + 1 + 2) = 48
(i) Probability of a black card (which are 26 – 3 =
23) $=\frac{23}{48}$
(ii) Probability of a being (4 – 2 = 2) = $\frac{1}{24}$
(iii) Probability of a red queen = (2 – 1 = 1) = $\frac{1}{48}$
Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.
Sol :
Total possible outcomes are . HH, HT, TT, TH, i.e., 4
(i) Favourable outcomes are HH, i.e., 1
So, P(2 heads)
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{1}{4}$
(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail)$=\frac{3}{4}$
Two different coins are tossed simultaneously. Find the probability of
getting :
(i) two tails
(ii) one tail
(iii) no tail
(iv) atmost one tail.
Sol :
Two different coins are tossed simultaneously
Number of possible outcomes = (2)² =4
Number of event having two tails = 1 i.e. (T, T)
(i) Probability of two tails will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{1}{4}$
(ii) Number of events having one tail = 2 i.e. (TH) and (HT)
Probability of one tail will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{1}{4}$
(iii) Number of events having no tail = 1 i.e. (HH)
Probability of having no tail will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{1}{4}$
(iv) Atmost one tail
Number Of events having at the most one tail = 3 i.e. (TH), (HT, (TT)
Probability of at the most one tail will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{3}{4}$
Two different dice are thrown simultaneously. Find the probability of
getting:
(i) a number greater than 3 on each dice
(ii) an odd number on both dice.
Sol :
When two different dice are thrown simultaneously,
then the sample space S of the random experiment =
{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) .
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
It consists of 36 equally likely outcomes.
(i) Let E be the event of ‘a number greater than 3 on each dice’.
E = {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
No. of favourable outcomes (E) = 9
P (number greater than 3 on each dice) $=\frac{9}{36}=\frac{1}{4}$
(ii) Let E be the event of ‘an odd number on both dice’.
E = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}
No. of favourable outcomes (E) = 9
∴ P (Odd on both dices) $=\frac{9}{36}=\frac{1}{4}$
Two different dice are thrown at the same time. Find the probability of
getting :
(i) a doublet
(ii) a sum of 8
(iii) sum divisble by 5
(iv) sum of atleast 11.
Sol :
Two different dice are thrown at the same time
Possible outcomes will be (6)² i.e. 36
(i) Number of events which doublet = 6
i.e. (1, 1), (2, 2) (3, 3), (4, 4), (5, 5) and (6, 6)
∴ Probability of doublets will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{6}{36}$
$=\frac{1}{6}$
(ii) Number of event in which the sum is 8 are
(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) = 5
Probability of a sum of 8 will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{5}{36}$
(iii) Number of event when sum is divisible by
5 are (1, 4), (4, 1), (2, 3), (3, 2), (4, 6),
(5, 5) = 7 in numbers
Probability of sum divisible by 5 will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{7}{36}$
(iv) Sum of atleast 11, will be in following events
(5, 6), (6, 5), (6, 6)
Probability of sum of atleast 11 will be
$P(E)=\frac{\text { Number } \text { of } \text { favourable } \text {
outcome }}{\text { Number of possible outcome }}$
$=\frac{3}{36}$
$=\frac{1}{12}$
