**Important questions for Class 10 Maths Chapter 15** **Probability** are given here based on the weightage prescribed by CBSE. The questions are framed as per the revised CBSE 2020-21 Syllabus and latest exam pattern. Students who are preparing for the CBSE class 10 board exams are advised to go through these questions of Probability to get the full marks for the questions from this chapter.

Students can also refer to the solutions prepared by BYJUâ€™S experts for all the chapters of Maths. Important questions for class 10 maths all chaptersÂ are also available to help the students in their examination preparation. The more students will practice, the more they can score marks in the exam. Along with the Important Question of class 10 Maths Chapter 15 Probability, students will also find the details solutions and answers. So, in case students could not solve any question, they can refer to the solution and can understand it easily.

Read more:

- Important 2 Marks Questions for CBSE 10th Maths
- Important 3 Marks Questions for CBSE 10th Maths
- Important 4 Marks Questions for CBSE 10th Maths

## Important Questions & Answers For Class 10 Maths Chapter 15 Probability

**Q. 1: Two dice are thrown at the same time. Find the probability of getting**

**(i) the same number on both dice.**

**(ii) different numbers on both dice.**

**Solution:**

Given that, Two dice are thrown at the same time.

So, the total number of possible outcomes n(S) = 6^{2} = 36

(i) Getting the same number on both dice:

Let A be the event of getting the same number on both dice.

Possible outcomes are (1,1), (2,2), (3, 3), (4, 4), (5, 5) and (6, 6).

Number of possible outcomes = n(A) = 6

Hence, the required probability =P(A) = n(A)/n(S)

= 6/36

= 1/6

(ii) Getting a different number on both dice.

Let B be the event of getting a different number on both dice.

Number of possible outcomes n(B) = 36 â€“ Number of possible outcomes for the same number on both dice

= 36 – 6 = 30

Hence, the required probability = P(B) = n(B)/n(S)

= 30/36

= 5/6

**Q. 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls beingÂ ****of the same size. Kritika takes out a ball from the bag without looking into it. What isÂ ****the probability that she takes out the**

**(i) yellow ball? **

**(ii) red ball? **

**(iii) blue ball?**

**Solution:**

Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them from the bag.

Let Y be the event â€˜the ball taken out is yellowâ€™, B be the event â€˜the ball taken out is blueâ€™, and R be the event â€˜the ball taken out is redâ€™.

The number of possible outcomes = Number of balls in the bag = n(S) = 3.

(i) The number of outcomes favourable to the event Y = n(Y) = 1.

So, P(Y) = n(Y)/n(S) =1/3

Similarly, (ii) P(R) = 1/3

and (iii) P(B) = â…“

**Q.3: One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will**

**(i) be an ace,**

**(ii) not be an ace.**

**Solution:**

Well-shuffling ensures equally likely outcomes.

(i) Card drawn is an ace

There are 4 aces in a deck.

Let E be the event â€˜the card is an aceâ€™.

The number of outcomes favourable to E = n(E) = 4

The number of possible outcomes = Total number of cards = n(S) = 52

Therefore, P(E) = n(E)/n(S) = 4/52 = 1/13

(ii) Card drawn is not an ace

Let F be the event â€˜card drawn is not an aceâ€™.

The number of outcomes favourable to the event F = n(F) = 52 â€“ 4 = 48

Therefore, P(F) = n(F)/n(S) = 48/52 = 12/13

**Q.4: Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown, and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.**

**Solution:**

Number of total outcome = n(S) = 36

(i) Let E_{1} be the event â€˜getting sum 2â€™

Favourable outcomes for the event E_{1} = {(1,1),(1,1)}

n(E_{1}) = 2

P(E1) = n(E1)/n(S) = 2/36 = 1/18

(ii) Let E_{2} be the event â€˜getting sum 3â€™

Favourable outcomes for the event E_{2} = {(1,2),(1,2),(2,1),(2,1)}

n(E_{2}) = 4

P(E_{2}) = n(E_{2})/n(S) = 4/36 = 1/9

(iii) Let E_{3} be the event â€˜getting sum 4â€™

Favourable outcomes for the event E_{3} = {(2,2)(2,2),(3,1),(3,1),(1,3),(1,3)}

n(E_{3}) = 6

P(E_{3}) = n(E_{3})/n(S) = 6/36 = 1/6

(iv) Let E_{4} be the event â€˜getting sum 5â€™

Favourable outcomes for the event E_{4} = {(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}

n(E_{4}) = 6

P(E_{4}) = n(E_{4})/n(S) = 6/36 = 1/6

(v) Let E_{5} be the event â€˜getting sum 6â€™

Favourable outcomes for the event E_{5} = {(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}

n(E_{5}) = 6

P(E_{5}) = n(E_{5})/n(S) = 6/36 = 1/6

(vi) Let E_{6} be the event â€˜getting sum 7â€™

Favourable outcomes for the event E_{6} = {(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}

n(E_{6}) = 6

P(E_{6}) = n(E_{6})/n(S) = 6/36 = 1/6

(vii) Let E_{7} be the event â€˜getting sum 8â€™

Favourable outcomes for the event E_{7} = {(5,3),(5,3),(6,2),(6,2)}

n(E_{7}) = 4

P(E_{7}) = n(E_{7})/n(S) = 4/36 = 1/9

(viii) Let E_{8} be the event â€˜getting sum 9â€™

Favourable outcomes for the event E_{8} = {(6,3),(6,3)}

n(E_{8}) = 2

P(E_{8}) = n(E_{8})/n(S) = 2/36 = 1/18

**Q.5: **A coin is tossed two times. Find the probability of getting at most one head.

**Solution:**

When two coins are tossed, the total no of outcomes = 2^{2} = 4

i.e. (H, H) (H, T), (T, H), (T, T)

Where,

H represents head

T represents the tail

We need at most one head, that means we need one head only otherwise no head.

Possible outcomes = (H, T), (T, H), (T, T)

Number of possible outcomes = 3

Hence, the required probability = Â¾

**Q.6: 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: **

Number of integers between 0 and 100 = n(S) = 99

(i) Let E be the event â€˜integer divisible by 7â€™

Favourable outcomes to the event E = 7, 14, 21,…., 98

Number of favourable outcomes = n(E) = 14

Probability = P(E) = n(E)/n(S) = 14/99

(ii) Let F be the event â€˜integer not divisible by 7â€™

Number of favourable outcomes to the event F = 99 – Number of integers divisible by 7

= 99-14 = 85

Hence, the required probability = P(F) = n(F)/n(S) = 85/99

**Q. 7:** **If P(E) = 0.05, what is the probability of â€˜not Eâ€™?**

**Solution:**

We know that,

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

It is given that, P(E) = 0.05

So, P(not E) = 1 – P(E)

P(not E) = 1 – 0.05

âˆ´ P(not E) = 0.95

**Q. 8:** **12 defective pens are accidentally mixed with 132 good ones. It is not possible to justÂ ****look at a pen and tell whether or not it is defective. One pen is taken out at random fromÂ ****this lot. Determine the probability that the pen is taken out is a good one.**

**Solution: **

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

âˆ´ 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

**Q. 9:** **A die is thrown twice. What is the probability that**

**(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]

**Solution:**

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 Ã— 6 = 36

(i) Method 1:

Consider the following events.

A = 5 comes in the first throw,

B = 5 comes in second throw

P(A) = 6/36,

P(B) = 6/36 and

P(not B) = 5/6

So, P(notA) = 1 – 6/36 = 5/6

âˆ´ The required probability = 5/6 Ã— 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 â€“ (5 + 6)] = 25

âˆ´ P(E) = 25/36

(ii) Number of events when 5 comes at least once = 11 (5 + 6)

âˆ´ The required probability = 11/36

**Q. 10:** **The probability that a non-leap year selected at random will contain 53 Sundays is**

**1/7****2/7****3/7****5/7**

**Solution: **

Correct option: (A)

A non-leap year has 365 days.

365 days = 52 weeks + 1 day

Since 52 weeks = 52 x 7 = 364 days

Hence there will be 52 Sundays in a year.

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday.

Out of these total 7 outcomes, the number of favourable outcomes is 1.

Hence, the probability of getting 53 Sundays = 1 / 7

### Practice Questions for Class 10 Maths Chapter 15 Probability

- A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.
- Find the probability of getting a number less than 5 in a single throw of a die.
- A card is drawn from an ordinary pack and a gambler bets that it is a spade or an ace. What are the odds against his winning this bet?
- 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 be double than that in case (i). Find x.
- Five male and three female candidates are available for selection as a manager in a company. Find the probability that a male candidate is selected.
- A box contains cards numbered 6 to 50. A card is drawn at random from the box. Calculate the probability that the drawn card has a number which is a perfect square.
- In a single throw of a pair of different dice, what is the probability of getting

(i) a prime number on each dice?

(ii) a total of 9 or 11?

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