ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22
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- ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test
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 ?
Solution:
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 540 }{ 600 } \)
= \(\\ \frac { 9 }{ 10 } \)
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.
Solution:
Number of prized tickets = 5
Number of blank tickets = 995
Total number of tickets = 5 + 995 = 1000
Probability of prized ticket
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 1000 } \)
= \(\\ \frac { 1 }{ 200 } \)
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.
Solution:
Number of defective pens = 12
Number of good pens = 132
Total number of pens =12 + 132 = 144
Probability of good pen
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 132 }{ 144 } \)
= \(\\ \frac { 11 }{ 12 } \)
Question 5.
If the probability of winning a game is \(\\ \frac { 5 }{ 11 } \), what is the probability of losing ?
Solution:
Probability of winning game = \(\\ \frac { 5 }{ 11 } \)
⇒ P(E) = \(\\ \frac { 5 }{ 11 } \)
We know that P (E) + P (\(\overline { E } \)) = 1
where P (E) is the probability of losing the game.
\(\\ \frac { 5 }{ 11 } \) + P (\(\overline { E } \)) = 1
⇒ P (\(\overline { E } \)) = \(1- \frac { 5 }{ 11 } \)
= \(\\ \frac { 11-5 }{ 11 } \)
= \(\\ \frac { 6 }{ 11 } \)
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 ?
Solution:
Probability of Sania’s winning the game = 0.69
Let P (E) be the probability of Sania’s winning the game
and P (\(\overline { E } \)) be the probability of Sania’s losing
the game or probability of Sonali, winning the game
P (E) + P (\(\overline { E } \)) = 1
⇒ 0.69 + P (\(\overline { E } \)) = 1
⇒ P(\(\overline { E } \)) = 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 ?
Solution:
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 (\(\overline { E } \)) will be the probability of not red balls.
P (E) + P (\(\overline { E } \)) = 1
(i) But P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 8 } \)
(ii) P (\(\overline { E } \)) = 1 – P(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 ?
Solution:
Number of total students = 40
Number of girls = 25
Number of boys = 40 – 25 = 15
(i) Probability of a girl
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 25 }{ 40 } \)
= \(\\ \frac { 5 }{ 8 } \)
(ii) Probability of a boy
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 15 }{ 40 } \)
= \(\\ \frac { 3 }{ 8 } \)
Question 9.
A letter is chosen from the word ‘TRIANGLE’. What is the probability that it is a vowel ?
Solution:
There are three vowels: I, A, E
.’. The number of letters in the word ‘TRIANGLE’ = 8.
Probability of vowel
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 8 } \)
Question 10.
A letter of English alphabet is chosen at random. Determine the probability that the letter is a consonant.
Solution:
No. of English alphabet = 26
No. of vowel = 5
No. of constant = 25 – 5 = 21
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 21 }{ 26 } \)
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7 }{ 15 } \)
(ii) Probability of black or white balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5+3 }{ 15 } \)
= \(\\ \frac { 8 }{ 15 } \)
(iii) Probability of not black balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7+3 }{ 15 } \)
= \(\\ \frac { 10 }{ 15 } \)
= \(\\ \frac { 2 }{ 3 } \)
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?
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 20 } \)
= \(\\ \frac { 1 }{ 4 } \)
(ii) Probability (of a blue or black marble)
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7+5 }{ 20 } \)
= \(\\ \frac { 12 }{ 20 } \)
= \(\\ \frac { 3 }{ 5 } \)
(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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 22 } \)
= \(\\ \frac { 4 }{ 11 } \)
(ii) Probability of red or black balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+3 }{ 22 } \)
= \(\\ \frac { 9 }{ 22 } \)
(iii) Probability of not green balls i.e. having red, white and black balls.
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+8+3 }{ 22 } \)
= \(\\ \frac { 17 }{ 22 } \)
(iv) Probability of neither white nor black balls i.e. red and green balls
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6+5 }{ 22 } \)
= \(\\ \frac { 11 }{ 22 } \)
= \(\\ \frac { 1 }{ 2 } \)
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?
Solution:
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 ?
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 88 }{ 100 } \)
= \(\\ \frac { 22 }{ 25 } \)
(ii) Probability of shirts acceptable to Salim
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 88+8 }{ 100 } \)
= \(\\ \frac { 96 }{ 100 } \)
= \(\\ \frac { 24 }{ 25 } \)
Question 16.
A die is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2 ?
Solution:
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 6 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ii) A number less them 5 are 1, 2, 3, 4
Probability of a number less than 5 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 6 } \)
= \(\\ \frac { 2 }{ 3 } \)
(iii) A number greater than 5 is 6
Probability of a number greater than 5 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 6 } \)
(iv) Prime number is 2, 3, 5
Probability of a prime number is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 6 } \)
= \(\\ \frac { 1 }{ 2 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 6 } \)
= \(\\ \frac { 1 }{ 3 } \)
(vii) Numbers between 3 and 6 is 4, 5
Probability of a number between 3 and 6 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 6 } \)
= \(\\ \frac { 1 }{ 3 } \)
(viii) Numbers divisible by 2 or 3 are 2, 4 or 3,
Probability of a number between 2 or 3 is
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 6 } \)
= \(\\ \frac { 1 }{ 3 } \)
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 ?
Solution:
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 = \(\\ \frac { n(E) }{ n(S) } \)
= \(\\ \frac { 1 }{ 6 } \)
Question 19.
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?
Solution:
On the face of a game, numbers 1 to 8 is shown.
Possible outcomes = 8
(i) Probability of number 8 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 8 } \)
(ii) Odd number are 1, 3, 5, 7
Probability of a number which is an odd will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 8 } \)
= \(\\ \frac { 1 }{ 2 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 8 } \)
= \(\\ \frac { 3 }{ 4 } \)
(iv) A number less than 9 is 8.
Probability of a number less than 9 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 8 } \)
Question 20.
Find the probability that the month of January may have 5 Mondays in
(i) a leap year
(ii) a non-leap year.
Solution:
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 = \(\\ \frac { 3 }{ 7 } \)
Question 21.
Find the probability that the month of February may have 5 Wednesdays in
(i) a leap year
(ii) a non-leap year.
Solution:
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
Question 22.
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.
Solution:
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) = \(\\ \frac { n(N) }{ n(S) } \) = \(\\ \frac { 10 }{ 16 } \) = \(\\ \frac { 5 }{ 8 } \)
Question 23.
An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?
Solution:
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 = \(\\ \frac { 85 }{ 99 } \)
Question 24.
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)
Solution:
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) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 8 }{ 20 } \)
= \(\\ \frac { 2 }{ 5 } \)
(ii) Numbers divisible by 3 are 3, 6, 9, 12, 15, 18
No. of actual events = 6
P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 6 }{ 20 } \)
= \(\\ \frac { 3 }{ 10 } \)
(iii) Numbers which are perfect squares = 1, 4, 9, 16 = 4
P(E) = \(\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events } \)
= \(\\ \frac { 4 }{ 20 } \)
= \(\\ \frac { 1 }{ 5 } \)
Question 25.
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
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 25 } \)
(iii) Multiples of 6 are 6, 12, 18, 24
Number of multiples = 4
Probability of multiples of 6 will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 25 } \)
Question 26.
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 15 } \)
(ii) Prime number are 2, 3, 5, 7, 11, 13
Number of primes is 6
Probability of prime number will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 15 } \)
= \(\\ \frac { 2 }{ 5 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 15 } \)
= \(\\ \frac { 1 }{ 3 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 10 }{ 15 } \)
= \(\\ \frac { 2 }{ 3 } \)
(v) Perfect squares number are 1, 4, 9 i.e., 3 number
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 15 } \)
= \(\\ \frac { 1 }{ 5 } \)
Question 27.
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 19 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 19 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 16 }{ 19 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 19 } \)
Question 28.
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.
Solution:
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 = \(\\ \frac { 4 }{ 48 } \)
= \(\\ \frac { 1 }{ 12 } \)
Question 29.
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 14 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7 }{ 14 } \)
= \(\\ \frac { 1 }{ 2 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 14 } \)
Question 30.
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.
Solution:
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
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 81 }{ 90 } \)
= \(\\ \frac { 9 }{ 10 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 90 } \)
= \(\\ \frac { 1 }{ 10 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 18 }{ 90 } \)
= \(\\ \frac { 1 }{ 5 } \)
Question 31.
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 50 }{ 100 } \)
= \(\\ \frac { 1 }{ 2 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 12 }{ 100 } \)
= \(\\ \frac { 3 }{ 25 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 9 }{ 100 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 10 }{ 100 } \)
= \(\\ \frac { 1 }{ 10 } \)
Question 32.
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.
Solution:
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
\(2\times \frac { 15-x }{ 15 } =\frac { x }{ 15 } \)
⇒ 2(15 – x) = x
⇒ 30 – 2x = x
⇒ 30 = x + 2x
⇒ 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.
Solution:
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
⇒ \(\frac { x }{ x+6 } =2\times \frac { 6 }{ x+6 } \)
⇒ \(\frac { x }{ x+6 } =\frac { 12 }{ x+6 } \)
⇒ x = 12
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.
Solution:
In a bag, there are 24 balls
Since, there are x balls red, 2 × balls white and 3 × balls blue
x + 2x + 3x = 24
Question 35.
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
Solution:
In a playing card, there are 52 cards
Number of possible outcome = 52
(i) Probability of‘2’ of spade will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 52 } \)
(ii) There are 4 jack card Probability of jack will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 4 }{ 52 } \)
= \(\\ \frac { 1 }{ 13 } \)
(iii) King of red colour are 2 in number
Probability of red colour king will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 2 }{ 52 } \)
= \(\\ \frac { 1 }{ 26 } \)
(iv) Cards of diamonds are 13 in number
Probability of diamonds card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 13 }{ 52 } \)
= \(\\ \frac { 1 }{ 4 } \)
(v) Number of kings and queens = 4 + 4 = 8
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 8 }{ 52 } \)
= \(\\ \frac { 2 }{ 13 } \)
(vi) Non-face cards are = 52 – 3 × 4 = 52 – 12 = 40
Probability of non-face card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 40 }{ 52 } \)
= \(\\ \frac { 10 }{ 13 } \)
(vii) Black face cards are = 2 × 3 = 6
Probability of black face card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 52 } \)
= \(\\ \frac { 3 }{ 26 } \)
(viii) No. of black cards = 13 x 2 = 26
Probability of black card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 26 }{ 52 } \)
= \(\\ \frac { 1 }{ 2 } \)
(ix) Non-ace cards are 12 × 4 = 48
Probability of non-ace card will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 48 }{ 52 } \)
= \(\\ \frac { 12 }{ 13 } \)
(x) Non-face card of black colours are 10 × 2 = 20
Probability of non-face card of black colour will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 20 }{ 52 } \)
= \(\\ \frac { 5 }{ 13 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 36 }{ 52 } \)
= \(\\ \frac { 9 }{ 13 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 38 }{ 52 } \)
= \(\\ \frac { 19 }{ 26 } \)
Question 36.
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
Solution:
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) = \(\\ \frac { 2 }{ 49 } \)
Question 37.
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.
Solution:
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) = \(\\ \frac { 2 }{ 48 } \) = \(\\ \frac { 1 }{ 24 } \)
(iii) Probability of a red queen = (2 – 1 = 1) = \(\\ \frac { 1 }{ 48 } \)
Question 38.
Two coins are tossed once. Find the probability of getting:
(i) 2 heads
(ii) at least one tail.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail) = \(\\ \frac { 3 }{ 4 } \)
Question 39.
Two different coins are tossed simultaneously. Find the probability of getting :
(i) two tails
(ii) one tail
(iii) no tail
(iv) atmost one tail.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(ii) Number of events having one tail = 2 i.e. (TH) and (HT)
Probability of one tail will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(iii) Number of events having no tail = 1 i.e. (HH)
Probability of having no tail will be
P(E) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 1 }{ 4 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 4 } \)
Question 40.
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.
Solution:
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) = \(\\ \frac { 9 }{ 36 } \) = \(\\ \frac { 1 }{ 4 } \)
Question 41.
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.
Solution:
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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 6 }{ 36 } \)
= \(\\ \frac { 1 }{ 6 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 5 }{ 36 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 7 }{ 36 } \)
(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) = \(\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome } \)
= \(\\ \frac { 3 }{ 36 } \)
= \(\\ \frac { 1 }{ 12 } \)
We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Ex 22, drop a comment below and we will get back to you at the earliest.