# Chapter 15 – Probability Questions and Answers: NCERT Solutions for Class 10 Maths

Class 10 Maths NCERT book solutions for Chapter 15 - Probability Questions and Answers.

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Class 10 Maths NCERT book solutions for Chapter 15 - Probability Questions and Answers.

(i) Probability of an event E + Probability of the event ‘not E’ = ___________.

(ii) The probability of an event that cannot happen is __________. Such an event is called ________.

(iii) The probability of an event that is certain to happen is _________. Such an event is called _________.

(iv) The sum of the probabilities of all the elementary events of an experiment is __________.

(v) The probability of an event is greater than or equal to ___ and less than or equal to __________.

(i) Probability of an event E + Probability of the event ‘not E’ =1.

(ii) The probability of an event that cannot happen is0. Such an event is calledan impossible event.

(iii) The probability of an event that is certain to happen is1. Such an event is calleda sure or certain event.

(iv) The sum of the probabilities of all the elementary events of an experiment is1.

(v) The probability of an event is greater than or equal to0and less than or equal to1.

(i) A driver attempts to start a car. The car starts or does not start.

(ii) A player attempts to shoot a basketball. She/he shoots or misses the shot.

(iii) A trial is made to Solution: a true-false question. The

(iv) A baby is born. It is a boy or a girl.

Solution:

(i) This statement does not have equally likely outcomes as the car may or may not start depending upon various factors like fuel, etc.

(ii) Even this statement does not have equally likely outcomes as the player may shoot or miss the shot.

(iii) This statement has equally likely outcomes as it is known that the solution is either right or wrong.

(iv) This statement also has equally likely outcomes as it is known that the newly born baby can either be a boy or a girl.

Tossing of a coin is a fair way of deciding because the number of possible outcomes are only 2 i.e. either head or tail. Since these two outcomes are an equally likely outcome, tossing is unpredictable and is considered to be completely unbiased.

(A) 2/3 (B) -1.5 (C) 15% (D) 0.7

The probability of any event (E) always lies between 0 and 1 i.e. 0 ≤ P(E) ≤ 1. So, from the above options, option (B) -1.5 cannot be the probability of an event.

We know that,

P(E)+P(not E) = 1

It is given that, P(E) = 0.05

So, P(not E) = 1-P(E)

Or, P(not E) = 1-0.05

∴ P(not E) = 0.95

(i) an orange flavored candy?

(ii) a lemon flavored candy?

(i) We know that the bag only contains lemon-flavored candies.

So, The no. of orange flavored candies = 0

∴ The probability of taking out orange flavored candies = 0/1 = 0

(ii) As there are only lemon flavored candies, P(lemon flavored candies) = 1 (or 100%)

Let the event wherein 2 students having the same birthday be E

Given, P(E) = 0.992

We know,

P(E)+P(not E) = 1

Or, P(not E) = 1–0.992 = 0.008

∴ The probability that the 2 students have the same birthday is 0.008

(i) red?

(ii) not red?

The total number of balls = No. of red balls + No. of black balls

So, the total no. of balls = 5+3 = 8

We know that the probability of an event is the ratio between the no. of favourable outcomes and the total number of outcomes.

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Probability of drawing red balls = P (red balls) = (no. of red balls/total no. of balls) = 3/8

(ii) Probability of drawing black balls = P (black balls) = (no. of black balls/total no. of balls) = 5/8

(i) red?

(ii) white?

(iii) not green?

The Total no. of balls = 5+8+4 = 17

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of red balls = 5

P (red ball) = 5/17 = 0.29

P (white ball) = 8/17 = 0.47

(iii) Total number of green balls = 4

P (green ball) = 4/17 = 0.23

∴ P (not green) = 1-P(green ball) = 1-(4/7) = 0.77

(i) will be a 50 p coin?

(ii) will not be a ₹5 coin?

Total no. of coins = 100+50+20+10 = 180

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i)Total number of 50 p coin = 100

P (50 p coin) = 100/180 = 5/9 = 0.55

P (₹5 coin) = 10/180 = 1/18 = 0.055

∴ P (not ₹5 coin) = 1-P (₹5 coin) = 1-0.055 = 0.945

The total number of fish in the tank = 5+8 = 13

Total number of male fish = 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (male fish) = 5/13 = 0.38

(i) 8?

(ii) an odd number?

(iii) a number greater than 2?

(iv) a number less than 9?

Total number of possible outcomes = 8

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total number of favourable events (i.e. 8) = 1

∴ P (pointing at 8) = ⅛ = 0.125

(ii) Total number of odd numbers = 4 (1, 3, 5 and 7)

P (pointing at an odd number) = 4/8 = ½ = 0.5

(iii) Total numbers greater than 2 = 6 (3, 4, 5, 6, 7 and 8)

P (pointing at a number greater than 4) = 6/8 = ¾ = 0.75

(iv) Total numbers less than 9 = 8 (1, 2, 3, 4, 5, 6, 7, and 8)

P (pointing at a number less than 9) = 8/8 = 1

(i) a prime number;

(ii) a number lying between 2 and 6;

(iii) an odd number.

Total possible events when a dice is thrown = 6 (1, 2, 3, 4, 5, and 6)

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i)Total number of prime numbers = 3 (2, 3 and 5)

P (getting a prime number) = 3/6 = ½ = 0.5

P (getting a number between 2 and 6) = 3/6 = ½ = 0.5

(iii)Total number of odd numbers = 3 (1, 3 and 5)

P (getting an odd number) = 3/6 = ½ = 0.5

(i) a king of red colour

(ii) a face card

(iii) a red face card

(iv) the jack of hearts

(v) a spade

(vi) the queen of diamonds

Total number of possible outcomes = 52

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i)Total numbers of king of red colour = 2

P (getting a king of red colour) = 2/52 = 1/26 = 0.038

P (getting a face card) = 12/52 = 3/13 = 0.23

P (getting a king of red colour) = 6/52 = 3/26 = 0.11

(iv)Total numbers of jack of hearts = 1

P (getting a king of red colour) = 1/52 = 0.019

(v)Total numbers of king of spade = 13

P (getting a king of red colour) = 13/52 = ¼ = 0.25

(vi)Total numbers of queen of diamonds = 1

P (getting a king of red colour) = 1/52 = 0.019

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?

Solution:

Total numbers of cards = 5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i)Numbers of queen = 1

P (picking a queen) = ⅕ = 0.2

(ii)If the queen is drawn and put aside, the total numbers of cards left is (5-4) = 4

(a) Total numbers of ace = 1

P (picking an ace) = ¼ = 0.25

(b) Total numbers of queen = 0

P (picking a queen) = 0/4 = 0

Numbers of pens = Numbers of defective pens + Numbers of good pens

∴ Total number of pens = 132+12 = 144 pens

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P(picking a good pen) = 132/144 = 11/12 = 0.916

(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?

(i)Number of defective bulbs = 4

The total number of bulbs = 20

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ Probability of getting a defective bulb = P (defective bulb) = 4/20 = ⅕ = 0.2

So, the total number of events (or outcomes) = 19

Number of non-defective bulbs = 19-4 = 15

So, the probability that the bulb is not defective = 15/19 = 0.789

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5.

The total number of discs = 90

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i)Total number of discs having two digit numbers = 81

(Since 1 to 9 are single digit numbers and so, total 2 digit numbers are 90-9 = 81)

P (bearing a two-digit number) = 81/90 = 9/10 = 0.9

(ii)Total number of perfect square numbers = 9 (1, 4, 9, 16, 25, 36, 49, 64 and 81)

P (getting a perfect square number) = 9/90 = 1/10 = 0.1

P (getting a number divisible by 5) = 18/90 = ⅕ = 0.2

(i) A?

(ii) D?

The total number of possible outcomes (or events) = 6

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i)The total number of faces having A on it = 2

P (getting A) = 2/6 = ⅓ = 0.33

P (getting D) = ⅙ = 0.166

First, calculate the area of the rectangle and the area of the circle. Here, the area of the rectangle is the possible outcome and the area of the circle will be the favourable outcome.

So, the area of the rectangle = (3×2) m2= 6 m2

and,

The area of the circle = πr2= π(½)2m2= π/4 m2= 0.78

∴ The probability that die will land inside the circle = [(π/4)/6] = π/24 or, 0.78/6 = 0.13

(i) She will buy it?

(ii) She will not buy it?

The total numbers of outcomes i.e. pens = 144

Given, numbers of defective pens = 20

∴ The numbers of non defective pens = 144-20 = 124

P(E) = (Number of favourable outcomes/ Total number of outcomes)

(i) Total numbers events in which she will buy them = 124

So, P (buying) = 124/144 = 31/36 = 0.86

So, P (not buying) = 20/144 = 5/36 = 0.138

Solution:

If 2 dices are thrown, the possible events are:

(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)

So, the total numbers of events: 6×6 = 36

(i)It is given that to get the sum as 2, the probability is 1/36 as the only possible outcomes = (1,1)

For getting the sum as 3, the possible events (or outcomes) = E (sum 3) = (1,2) and (2,1)

So, P(sum 3) = 2/36

Similarly,

E (sum 4) = (1,3), (3,1), and (2,2)

So, P (sum 4) = 3/36

E (sum 5) = (1,4), (4,1), (2,3), and (3,2)

So, P (sum 5) = 4/36

E (sum 6) = (1,5), (5,1), (2,4), (4,2), and (3,3)

So, P (sum 6) = 5/36

E (sum 7) = (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

So, P (sum 7) = 6/36

E (sum 8) = (2,6), (6,2), (3,5), (5,3), and (4,4)

So, P (sum 8) = 5/36

E (sum 9) = (3,6), (6,3), (4,5), and (5,4)

So, P (sum 9) = 4/36

E (sum 10) = (4,6), (6,4), and (5,5)

So, P (sum 10) = 3/36

E (sum 11) = (5,6), and (6,5)

So, P (sum 11) = 2/36

E (sum 12) = (6,6)

So, P (sum 12) = 1/36

So, the table will be as:

Event:

Sum on 2 dice

2

3

4

5

6

7

8

9

10

11

12

Probability

1/36

2/36

3/36

4/36

5/36

6/36

5/36

4/36

3/36

2/36

1/36

(ii)The argument is not correct as it is already justified in (i) that the number of all possible outcomes is 36 and not 11.

The total number of outcomes = 8 (HHH, HHT, HTH, THH, TTH, HTT, THT, TTT)

Total outcomes in which Hanif will lose the game = 6 (HHT, HTH, THH, TTH, HTT, THT)

P (losing the game) = 6/8 = ¾ = 0.75

(i) 5 will not come up either time?

(ii) 5 will come up at least once?

[Hint : Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Outcomes are:

(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)

So, the total number of outcome = 6×6 = 36

(i)Method 1:

Consider the following events.

A = 5 comes in first throw,

B = 5 comes in second throw

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

So, P(not A) = 1-(6/36) = 5/6

∴ The required probability = (5/6)×(5/6) = 25/36

Method 2:

Let E be the event in which 5 does not come up either time.

So, the favourable outcomes are [36–(5+6)] = 25

∴ P(E) = 25/36

∴ The required probability = 11/36

(i) If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is 1/3

(ii) If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is 1/2

(i)All the possible events are (H,H); (H,T); (T,H) and (T,T)

So, P (getting two heads) = ¼

and, P (getting one of the each) = 2/4 = ½

∴ This statement is incorrect.

(ii)Since the two outcomes are equally likely, this statement is correct.

(i) the same day?

(ii) consecutive days?

(iii) different days?

Since there are 5 days and both can go to the shop in 5 ways each so,

The total number of possible outcomes = 5×5 = 25

(i)The number of favourable events = 5 (Tue., Tue.), (Wed., Wed.), (Thu., Thu.), (Fri., Fri.), (Sat., Sat.)

So, P (both visiting on the same day) = 5/25 = ⅕

So, P(both visiting on the consecutive days) = 8/25

(iii)P (both visiting on the different days) = 1-P (both visiting on the same day)

So, P (both visiting on the different days) = 1-(⅕) = ⅘

(i) even?

(ii) 6?

(iii) at least 6?

The table will be as follows:

+

1

2

2

3

3

6

1

2

3

3

4

4

7

2

3

4

4

5

5

8

2

3

4

4

5

5

8

3

4

5

5

6

6

9

3

4

5

5

6

6

9

6

7

8

8

9

9

12

So, the total number of outcomes = 6×6 = 36

(i)E (Even) = 18

P (Even) = 18/36 = ½

(ii)E (sum is 6) = 4

P (sum is 6) = 4/36 = 1/9

(iii)E (sum is atleast 6) = 15

P (sum is atleast 6) = 15/36 = 5/12

is double that of a red ball, determine the number of blue balls in the bag.

It is given that the total number of red balls = 5

Let the total number of blue balls = x

So, the total no. of balls = x+5

P(E) = (Number of favourable outcomes/ Total number of outcomes)

∴ P (drawing a blue ball) = [x/(x+5)] ——–(i)

Similarly,

P (drawing a red ball) = [5/(x+5)] ——–(i)

From equation (i) and (ii)

x = 10

So, the total number of blue balls = 10

box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now

double of what it was before. Find x

Total number of black balls = x

Total number of balls = 12

P(E) = (Number of favourable outcomes/ Total number of outcomes)

P (getting black balls) = x/12 ——————-(i)

Now, when 6 more black balls are added,

Total balls become = 18

∴ Total number of black balls = x+6

Now, P (getting black balls) = (x+6)/18 ——————-(ii)

It’s given that,the probability of drawing a black ball now is double of what it was before

(ii) = 2 × (i)

(x+6)/18 = 2 × (x/12)

x + 6 = 3x

2x = 6

∴ x = 3

random from the jar, the probability that it is green is ⅔. Find the number of blue balls

in the jar.

Total marbles = 24

Let the total green marbles = x

So, the total blue marbles = 24-x

P(getting green marble) = x/24

From the question, x/24 = ⅔

So, the total green marbles = 16

And, the total blue marbles = 24-16 = 8

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