## Selina Concise Mathematics Class 10 ICSE Solutions Probability

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Mathematics Chapter 25 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Mathematics for Class 10 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 25 Probability

### Probability Exercise 25(A) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A coin is tossed once. Find the probability of:
(i) getting a tail
(ii) not getting a tail
Solution:

Question 2.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:

Question 3.
In a single throw of a die, find the probability of getting a number:
(i) greater than 4.
(ii) less than or equal to 4.
(iii) not greater than 4.
Solution:

Question 4.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will not be an even number.
(iii) will be an odd number.
Solution:

Question 5.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will:
(i) be a black card.
(ii) not be a red card.
(iii) be a red card.
(iv) be a face card.
(v) be a face card of red colour.
Solution:
Total number of cards = 52
Total number of outcomes = P(s) = 52
There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and diamond are black. So, there are 26 red cards and 26 black cards.
(i) Number of black cards in a deck = 26
P(E) = favourable outcomes for the event of drawing a black card = 26

(iv) There are 52 cards in a deck of cards, and 12 of these cards are face cards (4 kings, 4 queens, and 4 jacks).
P(E) = 12

Question 6.
(i) If A and B are two complementary events then what is the relation between P(A) and P(B)?
(ii) If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?
Solution:
(i) Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.
P(A) + P(B) = 1
(ii) P(A) = 0.46
Let P(B) be the probability of not happening of event A
We know,
P(A) + P(B) = 1
P(B) = 1 – P(A)
P(B) = 1 – 0.46
P(B) = 0.54
Hence the probability of not happening of event A is 0.54

Question 7.
In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:
(i) winning of Geeta
(ii) not winning of Ritu
Solution:
(i) Winning of Geeta is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(winning of Geeta) = 1
P(winning of Geeta) = 1 – P(winning of Ritu)
P(winning of Geeta) = 1 – 0.73
P(winning of Geeta) = 0.27
(ii) Not winning of Ritu is a complementary event to winning of Ritu
Therefore,
P(winning of Ritu) + P(not winning of Ritu) = 1
P(not winning of Ritu) = 1 – P(winning of Ritu)
P(not winning of Ritu) = 1 – 0.73
P(not winning of Ritu) = 0.27

Question 8.
In a race between Mahesh and John, the probability that John will lose the race is 0.54. Find the probability of:
(i) winning of Mahesh
(ii) winning of John
Solution:
(i) But if John looses, Mahesh wins
Hence, probability of John losing the race = Probability of Mahesh winning the race since it is a race between these two only
Therefore, P(winning of Mahesh) = 0.54
(ii) P(winning of Mahesh) + P(winning of John) = 1
0.54 + P(winning of John) = 1
P(winning of John) = 1 – 0.54
P(winning of John) = 0.46

Question 9.
(i) Write the probability of a sure event
(ii) Write the probability of an event when impossible
(iii) For an event E, write a relation representing the range of values of P(E)
Solution:

The number of elements in ‘E’ can’t be less than ‘0’ i.e. negative and greater than the number of elements in S.

Question 10.
In a single throw of die, find the probability of getting:
(i) 5
(ii) 8
(iii) a number less than 8
(iv) a prime number
Solution:

(ii) There are only six possible outcomes in a single throw of a die. If we want to find probability of 8 to come up, then in that case number of possible or favourable outcome is 0 (zero)

Question 11.
A die is thrown once. Find the probability of getting:
(i) an even number
(ii) a number between 3 and 8
(iii) an even number or a multiple of 3
Solution:

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/5
(ii) 2.7
(iii) 43%
(iv) -0.6
(v) -3.2
(vi) 0.35
Solution:

Question 13.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball
(ii) a black ball
Solution:

Question 14.
A single letter is selected at random from the word ‘Probability’. Find the probability that it is a vowel.
Solution:

Question 15.

Solution:

### Probability Exercise 25(B) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number
(ii) a multiple of 3
(iii) an even number and a multiple of 3
(iv) an even number or a multiple of 3
Solution:

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10
Favorable number of events = n(E) = 7

Question 2.
Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:
(i) a multiple of 5
(ii) a multiple of 6
(iii) between 40 and 60
(iv) greater than 85
(v) less than 48
Solution:

Question 3.
From 25 identical cards, numbered 1, 2, 3, 4, 5, ……, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 3
(ii) 5
(iii) 3 and 5
(iv) 3 or 5
Solution:

Question 4.
A die is thrown once. Find the probability of getting a number:
(i) less than 3
(ii) greater than or equal to 4
(iii) less than 8
(iv) greater than 6
Solution:

Question 5.
A book contains 85 pages. A page is chosen at random. What is the probability that the sum of the digits on the page is 8?
Solution:

Question 6.
A pair of dice is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
Solution:

Question 7.
If two coins are tossed once, what is the probability of getting:
(iii) both heads or both tails.
Solution:

Question 8.
Two dice are rolled together. Find the probability of getting:
(i) a total of at least 10.
(ii) a multiple of 2 on one die and an odd number on the other die.
Solution:

Question 9.
A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card drawn is:
(i) a spade(v) Jack or queen
(ii) a red card(vi) ace and king
(iii) a face card(vii) a red and a king
(iv) 5 of heart or diamond(viii) a red or a king
Solution:

Question 10.
A bag contains 16 colored balls. Six are green, 7 are red and 3 are white. A ball is chosen, without looking into the bag. Find the probability that the ball chosen is:
(i) red(v) green or red
(ii) not red(vi) white or green
(iii) white(vii) green or red or white
(iv) not white
Solution:

Question 11.
A ball is drawn at random from a box containing 12 white, 16 red and 20 green balls. Determine the probability that the ball drawn is:
(i) white(iii) not green
(ii) red(iv) red or white
Solution:

Question 12.
A card is drawn from a pack of 52 cards. Find the probability that the card drawn is:
(i) a red card
(ii) a black card
(iv) an ace
(v) a black ace
(vi) ace of diamonds
(vii) not a club
(viii) a queen or a jack
Solution:

Question 13.
Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:
(i) a multiple of 4 or 6
(ii) a multiple of 3 and 5
(iii) a multiple of 3 or 5
Solution:

Question 14.
In a single throw of two dice, find the probability of:
(i) a doublet
(ii) a number less than 3 on each dice
(iii) an odd number as a sum
(iv) a total of at most 10
(v) an odd number on one dice and a number less than or equal to 4 on the other dice.
Solution:

### Probability Exercise 25(C) – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A bag contains 3 red balls, 4 blue balls and 1 yellow ball, all the balls being identical in shape and size. If a ball is taken out of the bag without looking into it; find the probability that the ball is:
(i) yellow
(ii) red
(iii) not yellow
(iv) neither yellow nor red
Solution:

Question 2.
A dice is thrown once. What is the probability of getting a number:
(i) greater than 2?
(ii) less than or equal to 2?
Solution:

Question 3.
From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn is:
(i) a face card
(ii) not a face card
(iii) a queen of black card
(iv) a card with number 5 or 6
(v) a card with number less than 8
(vi) a card with number between 2 and 9
Solution:

Question 4.
In a match between A and B:
(i) the probability of winning of A is 0.83. What is the probability of winning of B?
(ii) the probability of losing the match is 0.49 for B. What is the probability of winning of A?
Solution:

Question 5.
A and B are friends. Ignoring the leap year, find the probability that both friends will have:
(i) different birthdays?
(ii) the same birthday?
Solution:

Question 6.
A man tosses two different coins (one of Rs 2 and another of Rs 5) simultaneously. What is the probability that he gets:
Solution:

Question 7.
A box contains 7 red balls, 8 green balls and 5 white balls. A ball is drawn at random from the box. Find the probability that the ball is:
(i) white
(ii) neither red nor white.
Solution:

Question 8.
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
Solution:

Question 9.
In a musical chairs game, a person has been advised to stop playing the music at any time within 40 seconds after its start. What is the probability that the music will stop within the first 15 seconds?
Solution:

Question 10.
In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:
(i) it is acceptable to a trader who accepts only a good shirt?
(ii) it is acceptable to a trader who rejects only a shirt with major defects?
Solution:

Question 11.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 8
(ii) 13
(iii) less than or equal to 12
Solution:

Question 12.
Which of the following cannot be the probability of an event?
(i) 3/7
(ii) 0.82
(iii) 37%
(iv) -2.4
Solution:

Question 13.
If P(E) = 0.59; find P(not E)
Solution:
P(E) + P(not E) = 1
0.59 + P(not E) = 1
P(not E) = 1 – 0.59 = 0.41

Question 14.
A bag contains a certain number of red balls. A ball is drawn. Find the probability that the ball drawn is:
(i) black
(ii) red
Solution:

Question 15.
The probability that two boys do not have the same birthday is 0.897. What is the probability that the two boys have the same birthday?
Solution:
P(do not have the same birthday)+P(have same birthday) = 1
0.897 + P(have same birthday) = 1
P(have same birthday) = 1 – 0.897
P(have same birthday) = 0.103

Question 16.
A bag contains 10 red balls, 16 white balls and 8 green balls. A ball is drawn out of the bag at random. What is the probability that the ball drawn will be:
(i) not red?
(ii) neither red nor green?
(iii) white or green?
Solution:

Question 17.
A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:
(i) will be a Re 1 coin?
(ii) will not be a Rs 2 coin?
(iii) will neither be a Rs 5 coin nor be a Re 1 coin?
Solution:

Question 18.
A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12; as shown below.
If the outcomes are equally likely, find the probability that the pointer will point at:
(i) 6
(ii) an even number
(iii) a prime number
(iv) a number greater than 8
(v) a number less than or equal to 9
(vi) a number between 3 and 11

Solution:

Question 19.
One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:
(i) a queen of red color
(ii) a black face card
(iii) the jack or the queen of the hearts
(iv) a diamond
(v) a diamond or a spade
Solution:

Question 20.
From a deck of 52 cards, all the face cards are removed and then the remaining cards are shuffled. Now one card is drawn from the remaining deck. Find the probability that the card drawn is:
(i) a black card
(ii) 8 of red color
(iii) a king of black color
Solution:

Question 21.
Seven cards:- the eight, the nine, the ten, jack, queen, king and ace of diamonds are well shuffled. One card is then picked up at random.
(i) What is the probability that the card drawn is the eight or the king?
(ii) If the king is drawn and put aside, what is the probability that the second card picked up is:
a) an ace? b) a king?
Solution:

Question 22.
A box contains 150 bulbs out of which 15 are defective. It is not possible to just look at a bulb and tell whether or not it is defective. One bulb is taken out at random from this box. Calculate the probability that the bulb taken out is:
(i) a good one
(ii) a defective one
Solution:

Question 23.
(i) 4 defective pens are accidentally mixed with 16 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is drawn at random from the lot. What is the probability that the pen is defective?
(ii) Suppose the pen drawn in (i) is defective and is not replaced. Now one more pen is drawn at random from the rest. What is the probability that this pen is:
a) defective
b) not defective?
Solution:

Question 24.
A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:
(i) a perfect square number
(ii) a number divisible by 4
(iii) a number divisible by 5
(iv) a number divisible by 4 or 5
(v) a number divisible by 4 and 5
Solution:

Question 25.
A circle with diameter 20 cm is drawn somewhere on a rectangular piece of paper with length 40 cm and width 30 cm. This paper is kept horizontal on table top and a die, very small in size, is dropped on the rectangular paper without seeing towards it. If the die falls and lands on paper only, find the probability that it will fall and land:
(i) inside the circle
(ii) outside the circle
Solution:

Question 26.
Two dice (each bearing numbers 1 to 6) are rolled together. Find the probability that the sum of the numbers on the upper-most faces of two dice is:
(i) 4 or 5
(ii) 7, 8 or 9
(iii) between 5 and 8
(iv) more than 10
(v) less than 6
Solution:

Question 27.
Three coins are tossed together. Write all the possible outcomes. Now, find the probability of getting:
(iv) all tails
(v) at least one tail
Solution:

Question 28.
Two dice are thrown simultaneously. What is the probability that:
(i) 4 will not come up either time?
(ii) 4 will come up at least once?
Solution:

Question 29.
Cards marked with numbers 1, 2, 3 ……… 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
Solution:

Question 30.
Offices in Delhi are open for five days in a week (Monday to Friday). Two employees of an office remain absent for one day in the same particular week. Find the probability that they remain absent on:
(i) the same day
(ii) consecutive day
(iii) different days
Solution:

Question 31.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball; find the number of black balls in the box.
Solution:

Question 32.
From a pack of 52 playing cards, all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is
(i) A face card (King, Jack or Queen)
(ii) An even numbered red card?
Solution:

Question 33.
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:
Given that the die has 6 faces marked by the given numbers as below:

When a die is rolled, total number of possible outcomes – 6
(i) For getting a positive integer, the favourable outoomes are: 1, 2, 3
⇒ Number of favourable outcomes – 3
⇒ Required probability = $$\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$$

(ii) For getting an integer greater than -3, the favourable outcomes
are: -2,-1, 1, 2, 3
⇒ Number of favourable outcomes – 5
⇒ Required probability = $$\frac { 5 }{ 6 }$$

(iii) For getting a smallest integer, the favourable outoomes are: -3
⇒ Number of favourable outcomes = 1
⇒ Required probability = $$\frac { 1 }{ 6 }$$

Question 34.
A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball.
(iii)Neither a green ball nor a white ball
Solution:
Number of white balls = 5
Number of red balls = 6
Number of green balls = 9
∴ Total number of balls = 5 + 6 + 9 = 20

Question 35.
A game of numbers has cards marked with 11, 12, 13, ….., 40. A card is drawn at random. Find the probability that the number on the card drawn is:
(i) A perfect square
(ii) Divisible by 7.
Solution:
Total number of outcomes = 30
(i) The perfect squares from 11 to 40 are 16, 25 and 36. So, the number of possible outcomes = 3 Hence, the probability that the number on the card drawn is a perfect square
= $$=\quad \frac { Number\quad of\quad possible\quad outcomes }{ Total\quad number\quad of\quad outcomes } =\frac { 3 }{ 30 } =\frac { 1 }{ 10 }$$
(ii) Among the given numbers, 14, 21, 28 and 35 are divisible by 7. So, the number of possible outcomes = 4 Hence, the probability that the number on the card drawn is divisible by 7
= $$\frac{\text { Number of possible autoomes }}{\text { Total number of outoomes }}=\frac{4}{30}=\frac{2}{15}$$

Question 36.
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, Total number of all possible outcomes = 16
i. a, e, i and o are the vowels.
Number of favourable outcomes = 4
∴ Required Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{4}{16}=\frac{1}{4}$$

ii. Number of consonants = 16 – 4 (vowels) = 12
∴ Number of favourable outcomes = 12
∴ Required Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{12}{16}=\frac{3}{4}$$

iii. Median contains 6 letters.
∴ Number of favourable outcomes = 16 – 6 = 10
∴ Required Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{16}=\frac{5}{8}$$

Question 37.
A box contains a certain number of balls. On each of 60% balls, letter A is marked. On each of 30% balls, letter B is marked and on each of remaining balls, letter C is marked. A ball is drawn from the box at random. Find the probability that the ball drawn is:
i. marked C
ii. A or B
iii. neither B nor C
Solution:
A box contains,
60% balls, letter A is marked.
30% balls, letter B is marked.
10% balls, letter C is marked.
i. Total number of all possible outcomes = 100
Number of favourable outcomes = 10
∴ Required Probability = $$\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{10}{100}=\frac{1}{10}$$

ii. The probability that the ball drawn is marked A = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}=\frac{60}{100}=\frac{6}{10}$$ … (1)
The probability that the ball drawn is marked B = $$\frac{\text { Number of favou rable outcomes }}{\text { Total number of all possible outcomes }}=\frac{30}{100}=\frac{3}{10}$$ … (2)
∴ Required Probability = $$\frac{6}{10}+\frac{3}{10}=\frac{9}{10}$$
iii. The probability that the ball drawn is neither B nor C
= 1 – [P(B) + P(C)]
= 1 – $$\left[\frac{3}{10}+\frac{1}{10}\right]$$
= 1 – $$\frac{4}{10}$$
= $$\frac{6}{10}$$
= $$\frac{3}{5}$$

Question 38.
A box contains a certain number of balls. Some of these balls are marked A, some are marked B and the remaining are marked C. When a ball is drawn at random from the box P(A) = $$\frac{1}{3}$$ and P(B) = $$\frac{1}{4}$$. If there are 40 balls in the box which are marked C, find the number of balls in the box.
Solution:
P(C) = 1 – [P(A) + P(B)] = $$1-\left[\frac{1}{3}+\frac{1}{4}\right]=1-\frac{7}{12}=\frac{5}{12}$$
Probability = $$\frac{\text { Number of favourable outcomes }}{\text { Total number of all possible outcomes }}$$
Given that 40 balls in the box are marked C.
⇒ $$\frac{5}{12}=\frac{40}{\text { Total number of all possible outcomes }}$$
⇒ Total number of all possible outcomes = $$\frac{40 \times 12}{5}=96$$
∴ the number of balls in the box is 96.

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

Selina Publishers Concise Mathematics Class 8 ICSE Solutions Chapter 23 Probability

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 8 Mathematics Chapter 23 Probability. You can download the Selina Concise Mathematics ICSE Solutions for Class 8 with Free PDF download option. Selina Publishers Concise Mathematics for Class 8 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

### Probability Exercise 23 – Selina Concise Mathematics Class 8 ICSE Solutions

Question 1.
A die is thrown, find the probability of getting:
(i) a prime number
(ii) a number greater than 4
(iii) a number not greater than 4.
Solution:
A die has six numbers : 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

Question 2.
A coin is tossed. What is the probability of getting:
Solution:
On tossing a coin once,
Number of possible outcome = 2

Question 3.
A coin is tossed twice. Find the probability of getting:
(i) exactly one head (ii) exactly one tail
(iii) two tails (iv) two heads
Solution:

Question 4.
A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?
Solution:
Total no. of letters in the word ‘PENCIL = 6
Total Number of Consonant = ‘PNCL’ i.e. 4

Question 5.
A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:
(i) a red ball
(ii) not a red ball
(iii) a white ball.
Solution:
Total number of possible outcomes = 3

Question 6.
6. In a single throw of a die, find the probability of getting a number
(i) greater than 2
(ii) less than or equal to 2
(iii) not greater than 2.
Solution:
A die has six numbers = 1, 2, 3, 4, 5, 6
∴ Number of possible outcomes = 6

Question 7.
A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.
A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:
(i) a black ball.
(ii) a red ball.
(iii) a white ball.
(iv) not a red ball.
(v) not a black ball.
Solution:

Question 8.
In a single throw of a die, find the probability that the number:
(i) will be an even number.
(ii) will be an odd number.
(iii) will not be an even number.
Solution:

Question 9.
In a single throw of a die, find the probability of getting :
(i) 8
(ii) a number greater than 8
(iii) a number less than 8
Solution:
On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.
∴ Number of possible outcome = 6

Question 10.
Which of the following can not be the probability of an event?

Solution:
The probability of an event cannot be
(ii) 3.8 i.e. the probability of an even cannot exceed 1.
(iv) i.e. -0.8 and
(vi) -2/5, This is because probability of an even can never be less than 1.

Question 11.
A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:
(i) a white ball,
(ii) a black ball
Solution:

Question 12.
Three identical coins are tossed together. What is the probability of obtaining:
Solution:

Question 13.
A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?
Solution:

Question 14.
Two coins are tossed together. What is the probability of getting:
(ii) both heads or both tails.
Solution:

Question 15.
From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:
(i) 2 (ii) 3
(iii) 2 and 3 (iv) 2 or 3
Solution:

Question 16.
Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:
(i) 0
(ii) 12
(iii) less than 12
(iv) less than or equal to 12
Solution:

Question 17.
A die is thrown once. Find the probability of getting:
(i) a prime number
(ii) a number greater than 3
(iii) a number other than 3 and 5
(iv) a number less than 6
(v) a number greater than 6.
Solution:

Question 18.
Two coins are tossed together. Find the probability of getting:
(i) exactly one tail
Solution:

## ICSE Solutions for Class 10 Mathematics – Probability

Get ICSE Solutions for Class 10 Mathematics Chapter 20 Probability for ICSE Board Examinations on APlusTopper.com. We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. You can download the Class 10 Maths ICSE Textbook Solutions with Free PDF download option.

### Formulae

1. If all the outcomes of an experiment are equally likely and E is an event, then probability of event E, written by P (E), is given by
2. 0 ≤ P (E) ≤ 1
3. P (not E) = 1 – P (E)
4. P (E) = 1 – P (not E)
5. P (E) + P (not E) = 1
6. The sum of the probabilities of all the elementary events of an experiment = 1
7. The probability of a sure event = 1
8. The probability of an impossible event = 0.

### Concept Based Questions

Question 1. An unbiased dice is thrown. What is the probability of getting a number other than 4.

Question 2. Two dice are thrown simulata- neously. Find the probability of getting six as the product.

Question 3. If the probability of winning a 5 game is 5/11. What is the probability of losing?

Question 4. Find the probability of getting a tail in a throw of a coin.

Question 5. In a cricket match a batsman hits a boundary 6 times out of 30 balls he play’s. Find the probability that he did not hit the boundary?

Question 6. It is known that a bax of 600 electric bulbs contain 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?

Question 7. 1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Namita has purchased one lottery ticket, what is the probability of winning a prize?

Question 8. A coin is tossed 100 times with the following frequencies:
Head = 55, Tail = 45
find the probability for each event (i) head (ii) tail.

Question 9. Namita tossed a coin once. What is the probability of getting (i) Head (ii) tail?

Question 10. Two coins are tossed once. Find the probability of getting.

Question 11. 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.

Question 12. 1800 families with 2 children were selected randomly and the following data were recorded:

Question 13. A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that at the card drawn is neither a red card nor a qeen.

Question 14. A dice is thrown once. What is the probability that the
(i) number is even
(ii) number is greater than 2?

Question 15. A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box.

Question 16. From a pack of 52 playing cards all cards whose numbers are multiples, of 3 are removed. A card is now drawn at random.
What is the probability that the card drawn is:
(i) a face card (King, Jack or Queen)
(ii) an even numbered red card?

Question 17. One card is randomly drawn from a pack of 52 cards. Find the probability that:
(i) the drawn card is red.
(ii) the drawn card is an ace.
(iii) the drawn card is red and a king.
(iv) the drawn card is red or king.

Question 18. One card is drawn from a pack of 52 cards, each of the 52 cards being equall likely to be drawn. Find the probability that the card drawn is (i) An ace, (ii) red, (iii) either red or king, (iv) red and a king, (v) a face card, (vi) a red face card, (vii) ‘2’ of spade, (viii) ’10’ of a blacksuit.

Question 19. Two dice are thrown simulta-neously. Find the probability of getting:
(i) an even number as the stun, (ii) the sum as a prime number, (iii) a total of at least 10, (iv) a doublet of even number, (v) a multiple of 3 as the sum.

Question 20. Find the probability that leap year selected at random, will contain 53 Sundays.

## Probability Involving AND and OR

Let’s examine AND first:

In probability, an outcome is in event
“A and B” only when the outcome is in both event A and event B.
In Venn Diagrams, we learned that an element was in the intersection “A and B”, only when the element was in BOTH sets.
n(A and B) means the number of outcomes in both A and B.
n(S) means the total number of possible outcomes

Example:
A die is rolled. What is the probability that the number is even and less than 4?
Answer: Event A: Numbers on a die that are even: 2, 4, 6
Event B: Numbers on a die that are less than 4: 1, 2, 3
There is only one number (2) that is in both events A and B.
Total outcomes S: Numbers on a die: 1, 2, 3, 4, 5, 6 (total = 6)
Probability = 1/6

Let’s examine OR:
In probability, an outcome is in event “A or B” when the outcome is in either (or both) event A or event B.
In Venn Diagrams, we learned that an element was in the union “A or B”, when the element was in either or both sets.
The rule for OR takes into account those values that may get counted more than once when the probability is determined. Check out the example below.

Example:
A die is rolled. What is the probability that the number is even or less than 4?
Event A: Numbers on a die that are even: 2, 4, 6 P(A)=3/6
Event B: Numbers on a die that are less than 4: 1, 2, 3 P(B)=3/6
P(A and B) = 1/6 (see rule above)
Answer: Probability = P(A) + P(B) – P(A and B)
= 3/6 + 3/6 – 1/6 = 5/6

Notice in this problem that the number 2 appears in both event A and event B. If we did not subtract the P(A and B), the answer would be 1 – which we know is not true since the number 5 appears in neither event.

## Define Probability and How do you find the Probability of an Event

Words ‘chance’ probably, or most probably etc. shows uncertainty in our statements. The uncertainity of ‘probably’ etc. can be measured numerically by means of ‘probability‘.
Trial and Event :
An experiment is called a trial if it results in anyone of the possible outcomes and  all the possible outcomes are called events.
For  Example

1. Participation of player in the game to win a game, is a trial but winning or losing is an event.
2. Tossing of a fair coin is a trial and turning up head or tail are events.
3. Throwing of a dice is a trial and occurrence of number 1 or 2 or 3 or 4 or 5 or 6 are events.
4. Drawing a card from a pack of playing cards is a trial and getting  an ace or a queen is an event.

Favourable Events :
Those outcomes of a trial  in which a given event may happen, are called favourable cases for that event.
For Example –

1. If a coin is tossed then favourable cases of getting H is 1.
2. If a dice is thrown then favourable case for getting 1 or 2 or 3 or 4 or 5 or 6, is 1.
3. If two dice are thrown, then favourable cases of getting a sum of numbers as 9 are four i.e (4,5), (5,4),  (3,6),  (6,3).

Sample Space :
The set of all possible outcomes of a trial is  called  its  sample  space. It is generally denoted by S and each outcome of the trial is said to be a point of sample of S.
For example

1. If a die is thrown once, then its sample space
S = {1, 2, 3, 4, 5, 6}
2. If two coins are tossed together then its sample space
S = {HT, TH, HH, TT}.

Mathematical definition of probability
Let there are n exhaustive, mutually exclusive and equally likely cases  for an event A and m of those are favourable to it, then probability of happening of the event A is defined by the  ratio m/n which is denoted by P(A). Thus
P(A) =  $$\frac{m}{n}$$
=  $$\frac{{No.\,of\,favourable\,\,cases\,\,to\,\,A}}{{No.\,of\,exhaustive\,\,cases\,\,to\,\,A}}$$
Note : It is obvious that 0 £ m £ n. If an event A is certain to happen, then m = n thus P (A) = 1.
If A is impossible to happen then m = 0 and so P (A) = 0. Hence we conclude that
0 P (A) 1
Further, if $$\bar A$$ denotes negative of A i.e. event that A doesn’t happen, then for above cases m, n ; we shall have
P ($$\bar A$$) = $$\frac{{n – m}}{n} = 1 – \frac{m}{n}$$ = 1– P (A)
P (A) + P ($$\bar A$$) = 1
Playing Cards :
(i)   Total : 52 (26 red, 26 black)
(ii)  Four suits : Heart, Diamond, Spade, Club – 13 cards each
(iii) Court Cards : 12 (4 Kings, 4 queens, 4 jacks)
(iv) Honour Cards: 16 (4 aces, 4 kings, 4 queens, 4 jacks)

## Probability Example Problems with Solutions

Example 1:     Two dice are thrown at a time. Find the probability of the following –
(i)   these numbers shown are equal;
(ii)  the difference of numbers shown is 1.
Solution:     The sample space in a throw of two dice
s = {1, 2, 3, 4, 5, 6} ×{1, 2, 3, 4, 5, 6}.
total no. of cases n (s) = 6 × 6 = 36.
(i)   Here E1 = the event of showing equal number on both dice
= {(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6) }
∴    n (E1 ) = 6
∴    P (E1) = $$\frac{{n({E_1})}}{{n(s)}} = \frac{6}{{36}} = \frac{1}{6}$$
(ii)  Here E2 = the event of showing numbers whose difference is 1.
= {(1, 2) (2, 1) (2, 3) (3, 2) (3, 4) (4, 3) (4, 5) (5, 4) (5, 6) (6, 5)}
∴    n (E2) = 10 ∴ p (E2) = $$\frac{{n({E_2})}}{{n(s)}} = \frac{{10}}{{36}} = \frac{5}{{18}}$$

Example 2:     Three coins are tossed together –
(i)   Find the probability of getting exactly two heads,
(ii)  Find the probability of getting at least two tails.
Solution:     The sample space in tossing three coins
S = (H, T) × (H, T) × (H, T)
∴    Total no. of cases n (s) = 2 × 2 × 2 = 8
(i)   Here E1 = the event of getting exactly two heads
= {HHT, HTH, THH}
∴   n (E1) = 3     ∴ P (E1) =  $$\frac{{n({E_1})}}{{n(s)}} = \frac{3}{8}$$
(ii)  E2 = {HTT, THT, TTH, TTT}
∴    n (E2) = 4,
∴    P(E2) = $$\frac{{n({E_2})}}{{n(s)}} = \frac{4}{8} = \frac{1}{2}$$

Example 3:     Find the probability of throwing (a) 3, (b) an even number with an ordinary six faced die.
Solution:     (a)  There are 6 possible ways in which the die can fall and there is only one way of throwing 3.
∴    The required probability
= $$\frac{{Number\;of\;favourable\;outcomes}}{{Total\;number\;of\;possible\;outcomes}} = \frac{1}{6}$$
(b)  Total number of outcomes of throwing a die = 6.
Number of outcomes of falling even number i.e.  2, 4, 6 = 3.
The required probability = $$\frac{3}{6} = \frac{1}{2}$$

Example 4:     A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen.
Solution:     There are 26 red cards (including 2 red queens) and 2 more queens are there. Thus, we have to set aside 28 cards.
And, we have to draw 1 card out of the remaining (52 – 28) = 24 cards.
∴    Required probability = $$\frac{24}{52} = \frac{6}{13}$$

Example 5:     Find the probability of getting a number less than 5 in a single throw of a die.
Solution:     There are 4 numbers which are less than 5, i.e. 1, 2, 3 and 4.
Number of such favourable outcomes = 4.
∵  The number marked on all the faces of a die are 1, 2, 3, 4, 5 or 6
∴    Total number of possible outcomes = 6
∴    P(a number less than 5) = $$\frac{4}{6} = \frac{2}{3}$$

Example 6:     If the probability of winning a game is 0.3, what is the probability of lossing it ?
Solution:     Probability of winning a game = 0.3.
Probability of losing it = q (say).
⇒   0.3 + q = 1
⇒  q = 1 – 0.3
⇒   q = 0.7

Example 7:     Two coins are tossed simulataneously. Find the probability of getting
Solution:     Let H denotes head and T denotes tail.
∴    On tossing two coins simultaneously, all the possible outcomes are
(i)   The probability of getting two heads = P(HH)
= $$\frac{{Event\;of\;occurence\;of\;two\;heads}}{{Total\;number\;of\;possible\;outcomes}} = \frac{1}{4}$$
(ii)  The probability of getting at least one head
= P(HT or TH or HH)
= $$\frac{{Event\;of\;occurence\;of\;at\;least\;one\;head}}{{Total\;number\;of\;possible\;outcomes}} = \frac{3}{4}$$
(iii) The probability of getting no head = P(TT)
= $$\frac{{Event\;of\;occurence\;of\;no\;head}}{{Total\;number\;of\;possible\;outcomes}} = \frac{1}{4}$$

Example 8:     On tossing three coins at a time, find –
(i)   All possible outcomes.
Solution:     Let H denotes head and T denotes tail. On tossing three coins at a time,
(i)   All possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. These are the 8 possible outcomes.
(ii)  An event of occurence of 3 heads
= (HHH) = 1
An event of occurence of 2 heads
= {HHT, HTH, THH} = 3
An event of occurence of 1 head
= {HTT, THT, TTH} = 3
An event of occurence of O head = {TTT} =1
(iii) Now, probability of getting 3 heads = P (HHH)
= $$\frac{{Event\;of\;occurence\;of\;3\;heads}}{{Total\;number\;of\;possible\;outcomes}} = \frac{1}{8}$$
Simultaneously, probability of getting 2 heads
= P(HHT or THH or HTH)
= $$\frac{{Event\;of\;occurence\;of\;2\;heads}}{{Total\;number\;of\;possible\;outcomes}} = \frac{3}{8}$$
= P (HTT or THT or TTH)
= $$\frac{{Event\;of\;occurence\;of\;1\;head}}{{Total\;number\;of\;possible\;outcomes}} = \frac{3}{8}$$
Probability of getting no head = P (TTT)
= $$\frac{{Event\;of\;occurence\;of\;no\;head}}{{Total\;number\;of\;possible\;outcomes}} = \frac{1}{8}$$

Example 9:     One card is drawn from a well-shiffled deck of 52 cards. Find the probability of drawing:
(i)   an ace
(iii) ‘10’ of black suit
Solution:     (i)   There are 4 aces in deck.
∴    Number of such favourable outcomes = 4
∵  Total number of cards in deck = 52.
∴    Total number of possible outcomes = 52.
∴    P(an ace) = $$\frac{4}{52} = \frac{1}{13}$$
(ii)  Number of ‘2’ of spades = 1
Number of favourable outcomes = 1
Total number of possible outcomes = 52
∴    P(‘2’ of spades) = $$\frac{1}{52}$$
(iii) There are 2 ‘10’ of black suits (i.e. spade and club)
∴    Number of favourable outcomes = 2
Total number of possible outcomes = 52
∴    P(‘10’ of a black suit) = $$\frac{2}{52} = \frac{1}{26}$$

Example 10:     A bag contains 12 balls out of which x are white,
(i)   If one ball is drawn at random, what is the probability that it will be a white ball ?
(ii)  If 6 more white balls are put in the bag, the probability of drawing a white ball will double than that in (i). Find x.
Solution:     Random drawing of balls ensures equally likely outcomes
Total number of balls = 12
∴    Total number of possible outcomes = 12
Number of white balls = x
(i)   Out of total 12 outcomes, favourable outcomes = x
P(White ball) = $$\frac{{Number\;of\;favourable\;outcomes}}{{Total\;number\;of\;possible\;outcomes}}$$
=  $$\frac{x}{12}$$
(ii)  If 6 more white balls are put in the bag, then
Total number of white balls = x + 6
Total number of balls in the bag
= 12 + 6 = 18
P(White ball) = $$\frac{{Number\;of\;favourable\;outcomes}}{{Total\;number\;of\;possible\;outcomes}}$$
= $$\frac{x+6}{12+6}$$
According to the question,
Probability of drawing white ball in second case
= 2 × probability drawing of white ball in first case
⇒   $$\frac{{x + 6}}{{18}} = 2\left( {\frac{x}{{12}}} \right)$$
⇒   $$\frac{{x + 6}}{{18}} = \frac{x}{6}$$
⇒   6x + 36 = 18x
⇒     12x = 36
⇒   x = 3
Hence, number of white balls = 3

Example 11:     What is the probability that a leap year, selected at random will contain 53 Sundays ?
Solution:     Number of days in a leap year = 366 days
Now, 366 days = 52 weeks and 2 days
The remaining two days can be
(i)   Sunday and Monday
(ii)  Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v)  Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
For the leap year to contain 53 Sundays, last two days are either Sunday and Monday or Saturday and Sunday.
∴    Number of such favourable outcomes = 2
Total number of possible outcomes = 7
∴    P(a leap year contains 53 sundays) = $$\frac{2}{7}$$

Example 12:     Three unbiased coins are tossed together. Find the probability of getting :
Solution:     Elementary events associated to random experiment of tossing three coins are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
∴    Total number of elementary events = 8.
(i)   The event “Getting all heads” is said to occur, if the elementary event HHH occurs i.e. HHH is an outcome. Therefore,
∴  Favourable number of elementary events = 1
Hence, required probability = $$\frac{1}{8}$$
(ii)  The event “Getting two heads” will occur, if one of the elementary events HHT, THH, HTH occurs.
∴  Favourable number of elementary events = 3
Hence, required probability = $$\frac{3}{8}$$
(iii) The events of getting one head, when three coins are tossed together, occurs if one of the elementary events HTT, THT, TTH happens.
∴  Favourable number of elementary events = 3
Hence, required probability = $$\frac{3}{8}$$
(iv) If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event “Getting at least two heads” occurs.
∴  Favourable number of elementary events = 4
Hence, required probability = $$\frac{4}{8}\,\, = \,\,\frac{1}{2}$$

Example 13:     17 Cards numbered 1, 2, 3 … 17 are put in a box and mixed thoroughly. One person draws a card from the box. Find the probability that the number on the card is
(i)   Odd
(ii)  A prime
(iii) Divisible by 3
(iv) Divisible by 3 and 2 both.
Solution:     Out of 17 cards, in the box, one card can be drawn in 17 ways.
∴    Total number of elementary events = 17.
(i)   There 9 odd numbered cards, namely, 1, 3, 5, 7, 9, 11, 13, 15, 17. Out of these 9 cards one card can be drawn in 9 ways.
∴ Favourable number of elementary events = 9.
Hence, required probability = $$\frac{9}{17}$$
(ii)  There are 7 prime numbered cards, namely, 2, 3, 5, 7, 11, 13, 17. Out of these 7 cards one card can be chosen in 7 ways.
∴  Favourable number of elementary events = 7.
Hence, P (Getting a prime number) = $$\frac{7}{17}$$
(iii) Let A denote the event of getting a card bearing a number divisible by 3. Clearly, event A occurs if we get a card bearing one of the numbers 3, 6, 9, 12, 15.
∴  Favourable number of elementary events = 5.
Hence, P (Getting a card bearing a number divisible by 3) = $$\frac{5}{17}$$
(iv) If a number is divisible by both 3 and 2, then it is a multiple of 6. In cards bearing number 1, 2, 3 …, 17 there are only 2 cards which bear a number divisible by 3 and 2 both i.e. by 6. These cards bear numbers 6 and 12
∴  Favourable number of elementary events = 2
Hence, P (Getting a card bearing a number divisible by 3 and 2) = $$\frac{2}{17}$$

Example 14:     A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is
(i) Black            (ii) Red         (iii) Not green.
Solution:     Total number of balls in the bag
= 5 + 8 + 4 + 7 = 24
∴ Total number of elementary events = 24
(i)   There are 7 black balls in the bag.
∴ Favourable number of elementary events = 7
Hence, P (Getting  a black ball) = $$\frac{7}{24}$$
(ii)  There are 5 red balls in the bag.
∴ Favourable number of elementary events = 5
Hence, P (Getting a red ball) = $$\frac{5}{24}$$
(iii) There are 5 + 8 + 7 = 20 balls which are not green.
∴ Favourable number of elementary events = 20
Hence, P (No getting a green ball) = $$\frac{4}{8}\,\, = \,\,\frac{1}{2}$$

Example 15:     Find the probability that a number selected at random from the numbers 1 to 25 is not a prime number when each of the given number is equally likely to be selected.
Solution:     Total number (1, 2, 3, 4, … 25) = 25.
Out of 25 numbers prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23.
So, remaining not a prime number are 25 – 9 = 16
Total number of possible outcomes = 25
and number of favourable outcomes = 16
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (not a prime) = $$\frac{{16}}{{25}}$$

Example 16:     A piggy bank contains hundred 50 p coins, fifty Re 1 coins, twenty Rs 2 coins and ten Rs 5 coins. If it is equally likely that one of the coins will fall out 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 a Rs. 5 coin ?
Solution:     Number of 50 p coins = 100
Number of 1 Rs coins = 50
Number of 2 Rs coins = 20
Number of 5 Rs coins = 10

(i)   The number of favourable outcomes of 50 p coin to fall = 100
Total number of coins = 100 + 50 + 20 + 10 = 180
Total number of possible outcomes = 180
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (50 p) = $$\frac{{100}}{{180}} = \frac{5}{9}$$
(ii)  Number of favourable outcomes of 5 Rs coin to not fall = 180 – 10 = 170
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (not Rs. 5) = $$\frac{{170}}{{180}} = \frac{17}{18}$$

Example 17:     (i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective ?
(ii)  Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution:     (i)   The total number of bulbs = 20
Total number of possible outcomes = 20
Number of favourable outcomes of defective bulbs = 4
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$

P (defective bulb) = $$\frac{4}{{20}}\,\, = \,\,\frac{1}{5}$$
(ii)  The bulb drawn is not defective
Total number of bulbs without replacement = 19
Number of defective bulbs = 4
Number of non defective bulbs = 19 – 4 = 15
Number of favourable outcomes of non defective bulbs = 15
Total number of possible outcomes = 19
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (non defective) = $$\frac{{15}}{{19}}$$

Example 18:     A game of chance consists of spinning an arrow which comes to rest pointing at one of the number 1, 2, 3, 4, 5, 6, 7, 8 (see fig), 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:     Total number of possible outcomes in the game = 8
(i)   Number of rest of arrow on 8 = 1
Number of favourable outcomes of 8 = 1
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (8) = $$\frac{1}{8}$$
(ii)  In the game the number of odd number
1, 3, 5, 7 = 4
Number of favourable outcomes of odd number = 4
P = (odd number) = $$\frac{4}{8}\,\, = \,\,\frac{1}{2}$$
(iii) Numbers greater than 2 = 6
Number of favourable outcomes of greater than 2 = 6
P = (greater than 2) = $$\frac{6}{8}\,\, = \,\,\frac{3}{4}$$
(iv) Number less than 9 = 8
Number of favourable outcome of less than 9 = 8
P (less than 9) = $$\frac{8}{8}$$ = 1

Example 19:     It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday ?
Solution:     Probability of 2 students from a group of 3 students not having the same birthday = 0.992
Probability of 2 students from a group of 3 students having the same birthday
[∴  p + q = 1] = 1 – 0.992 = 0.008

Example 20:     A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability that the card drawn is neither a red card nor a queen.
Solution:     Number of red cards including 2 red queens = 26
Number of black queens = 2
Therefore, number of red cards including 2 red queens and 2 black queens = 26 + 2 = 28
Number of cards neither a red card nor a queen = 52 – 28 = 24
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (neither a red nor a queen card) = $$\frac{{24}}{{52}}\,\, = \,\,\frac{6}{{13}}$$

Example 21:     A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing (i) a face card                (ii) a red face card.
Solution:     Random drawing of cards ensures equally likely outcomes
(i)   Number of face cards (King, queen and Jack of each suits) = 3 × 4 = 12
Total number of cards in a deck = 52
∴ Total number of possible outcomes = 52
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (drawing a face card) = $$\frac{{12}}{{52}}\,\, = \,\,\frac{3}{{13}}$$
(ii)  Number of red face cards 2 × 3 = 6
Number of favourable outcomes of drawing red face card = 6
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (drawing of red face card) = $$\frac{{6}}{{52}}\,\, = \,\,\frac{3}{{26}}$$

Example 22:     Two dice are thrown simultaneously. Fill up the table for number of events of sum on two dice.

 Events: ‘sum on 2 dice’ Probability 2 $$\frac{1}{36}$$ 3 4 5 6 7 8 $$\frac{5}{36}$$ 9 10 11 12 $$\frac{1}{36}$$

Solution:     Total number of possible outcomes = 6 × 6 = 36
(i)   Number of favourable outcomes of sum (2) = (1,1) = 1
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
⇒   P (sum, 2) = $$\frac{1}{36}$$
(ii)  Number of favourable outcomes of sum (3) is (1, 2), (2, 1) = 2
P (sum, 3) = $$\frac{2}{36}$$
(iii) Favourable outcomes of sum (4) are {2, 2), (1, 3), (3, 1)}
Number of favourable outcomes of sum (4) = 3
P (sum, 4) = $$\frac{3}{36}$$
(iv) Favourable outcomes of sum (5) are {(1, 4), (4, 1), (2, 3), (3, 2)}
Number of favourable outcomes of sum (5) = 4
P (sum, 5) = $$\frac{4}{36}$$
(v)  Favourable outcomes of sum (6) are {(1, 5), (5, 1), (2, 4), (4, 2), (3, 3)}
Number of favourable outcomes of sum (6) = 5
P (sum, 6) = $$\frac{5}{36}$$
(vi) Favourable outcomes of sum (7) are {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
Number of favourable outcomes of sum (7) = 6
P (sum, 7) = $$\frac{6}{36}$$
(vii)      Favourable outcomes of sum (8) are {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}
Number of favourable outcomes of sum (8) = 5
P (sum, 8) = $$\frac{5}{36}$$
(viii) Favourable outcomes of sum (9) are {(3, 6), (6, 3), (4, 5), (5, 4)}
Number of favourable outcomes of sum (9) = 4  ; P (sum, 9) = $$\frac{4}{36}$$
(ix) Favourable outcomes of sum (10) are {(4, 6), (6, 4), (5, 5)}
Number of favourable outcomes of sum (10) = 3 ;  P (sum, 10) = $$\frac{3}{36}$$
(x)  Favourable outcomes of sum (11) are {(6, 5), (5, 6)}
Number of favourable outcomes of sum(11) = 2 ;  P (sum, 11) = $$\frac{2}{36}$$
(xi) Favourable outcomes of sum (12) are (6, 6)
Number of favourable outcomes of sum (12) = 1 ;  P (sum, 12) = $$\frac{1}{36}$$

 Events: ‘sum on 2 dice’ Probability 2 $$\frac{1}{36}$$ 3 $$\frac{2}{36}$$ 4 $$\frac{3}{36}$$ 5 $$\frac{4}{36}$$ 6 $$\frac{5}{36}$$ 7 $$\frac{6}{36}$$ 8 $$\frac{5}{36}$$ 9 $$\frac{4}{36}$$ 10 $$\frac{3}{36}$$ 11 $$\frac{2}{36}$$ 12 $$\frac{1}{36}$$

Example 23:     Two customers Abbas and Shehla are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any one day as on another. What is the probability that both will visit the shop on
(i)   the same day
(ii) different days
(iii) consecutive days ?
Solution:     Two customers Abbas and Shehla visiting a shop Tuesday to Saturday.
Total possible ways of visiting shop by them
= 5 × 5
= 25
(i)   They can visit the shop on all week days Tuesday to Saturday.
Favourable outcomes of visiting shop by them on the same day = 5
Probability = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (visiting shop same day)
= $$\frac{5}{{25}}\,\, = \,\,\frac{1}{5}$$
(ii)  Favourable outcomes of visiting shop on the different days by them
= 25 – 5
= 20 days
P = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
P (visiting shop different days)
= $$\frac{20}{{25}}\,\, = \,\,\frac{4}{5}$$
(iii) Favourable outcomes of visiting shop by them on consecutive days are

 Abbas T W Th F Shehla W Th F S
 Shehla T W Th F Abbas W Th F S

Total favourable outcomes = 4 + 4 = 8 days
Number of favourable outcomes = 8
P (visiting shop on consecutive days)
= $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{8}{{25}}$$

Example 24:     A box contains 12 balls out of which x are black.
(i)   If one ball is drawn at random from the box, what is the probability that it will be a black ball ?
(ii)  If 6 more white balls are put in the bag, the probability of drawing a black ball will double than that in (i). Find x.
Solution:     Random drawing of balls ensures equally likely outcomes
Total number of balls = 12
∴    Total number of possible outcomes = 12
Number of black balls = x
(i)   Out of total 12 outcomes, favourable outcomes = x
P (black ball)
= $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{x}{12}$$
(ii)  If 6 more black balls are put in the bag, then
Total number of black balls = x + 6
Total number of balls in the bag = 12 + 6 = 18
P (black ball) = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{x+6}{12+6}$$
According to the question,
Probability of drawing black ball in second case
= 2 × probability drawing of black ball in first case
⇒ $$\frac{{x + 6}}{{18}} = 2\left( {\frac{x}{{12}}} \right)$$
⇒ $$\frac{{x + 6}}{{18}} = \frac{x}{6}$$
⇒ 6x + 36 = 18x
⇒ 12x = 36
⇒ x = 3
Hence, number of black balls = 3

Example 25:     A box contains 20 balls bearing numbers, 1, 2, 3, 4, … 20. A ball is drawn at random from the box. What is the probability that the number on the balls is
(i) An odd number
(ii) Divisible by 2 or 3
(iii) Prime number
(iv) Not divisible by 10
Solution:     Total number of possible outcomes = 20
Probability = $$\frac{{Number\,\,of\,favourable\,\,outcomes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
(i)   Number of odds out of first 20 numbers = 10
Favourable outcomes by odd = 10
P(odds) = $$\frac{{Favourable\,\,outcomes\,\,of\,\,odd}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{{10}}{{20}} = \frac{1}{2}$$
(ii)  The numbers divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20.
Favourable outcomes of numbers divisible by 2 or 3 = 13
P (numbers divisible by 2 or 3)
= $$\frac{{Favourable\,\,outcomes\,\,of\,\,divisible\,\,by\,\,2\,\,or\,\,3}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{{13}}{{20}}$$
(iii) Prime numbers out of first 20 numbers are 2, 3, 5, 7, 11, 13, 17, 19
Favourable outcomes of primes = 8
P(primes)
= $$\frac{{Favourable\,\,outcomes\,\,of\,\,primes}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{8}{{20}}\,\, = \,\,\frac{2}{5}$$
(iv) Numbers not divisible by 10 are 1, 2, … 9, 11, …19
Favourable outcomes of not divisible by 10
= 18
P(not divisible by 10)
= $$\frac{{Favourable\,\,outcomes\,\,of\,\,not\,\,divisible\,\,by\,\,10}}{{Total\,\,number\,\,of\,possible\,\,outcomes}}$$
= $$\frac{{18}}{{20}}\,\, = \,\,\frac{9}{{10}}$$