**Important questions for class 11 Maths Chapter 16 Probability** are given here, which are taken from the previous year’s question paper. Go through all the problems provided here, which will help you to score good marks in Class 11 final examination. Before solving the problems, be thorough with the concepts covered in the chapter – Probability. All the important questions in chapter 16 probability are solved here in a step-by-step procedure. Students can prepare these questions for the upcoming examinations because most of the questions are asked from the previous year’s question papers. Also, get important questions for class 11 Maths for all the chapters at BYJU’S.

Class 11 Maths Chapter 16 – Probability covers important concepts, such as random experiment, algebra of events, events and types of events, axiomatic approach to the probability

**Also, check:**

- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths

## Important Questions for Class 11 Maths Chapter 16 Probability with Solutions

The most important questions for class 11 Maths Chapter 16 probability are provided with solutions. It includes MCQs, short type and long type answers. Practice these problems in a smart way a achieve good marks in the examination.

**Question 1:**

If P(A) is ⅗. Find P (not A)

**Solution:**

Given that: P(A) = ⅗

To find P(not A) = 1 – P(A)

P (not A) = 1- ⅗

= (5-3)/5

= ⅖

Therefore, P(not A) = ⅖.

**Question 2:**

Find the probability that when a hand of 7 cards are drawn from the well-shuffled deck of 52 cards, it contains

(i) all kings and (ii) 3 kings

**Solution:**

**(i) To find the probability that all the cards are kings:**

If 7 cards are chosen from the pack of 52 cards

Then the total number of combinations possible is: ^{52}C_{7}

= 52!/[7! (52-7)!]

= 52!/ (7! 45!)

Assume that A be the event that all the kings are selected

We know that there are only 4 kings in the pack of 52 cards

Thus, if 7 cards are chosen, 4 kings are chosen out of 4, and 3 should be chosen from the 48 remaining cards.

Therefore, the total number of combinations is:

n(A) = ^{4}C_{4 } x^{48}C_{3}

= [4!/4!0! ] x [48!/3!(48-3)!]

= 1 x [48!/3! 45!]

= 48!/3! 45!

Therefore, P(A) = n(A)/n(S)

= [48!/3! 45!] ÷[52!/ (7! 45!]

= [48! x 7!] ÷ [3!x 52!]

= 1/7735

Therefore, the probability of getting all the 7 cards are kings is 1/7735

**(ii) To find the probability that 3 cards are kings:**

Assume that B be the event that 3 kings are selected.

Thus, if 7 cards are chosen, 3 kings are chosen out of 4, and 4 cards should be chosen from the 48 remaining cards.

Therefore, the total number of combinations is:

n(B) = ^{4}C_{3 } x^{48}C_{4}

= [4!/3!(4-3)! ] x [48!/4!(48-4)!]

= 4 x [48!/4! 44!]

= 48!/3! 45!

Therefore, P(B) = n(B)/n(S)

= [4 x48!/4! 44!] ÷[52!/ (7! 45!]

= 9/1547

Therefore, the probability of getting 3 kings is 9/1547

**Question 3: **

An urn contains 6 balls of which two are red and four are black. Two balls

are drawn at random. The probability that they are of different colours is

(i)⅖ (ii)1/15 (iii)8/15 (iv)4/15

**Solution:**

A correct answer is** an option (c)**

**Explanation:**

Given that, the total number of balls = 6 balls

Let A and B be the red and black balls respectively,

The probability that two balls drawn, are different = P(the first ball drawn is red)(the second ball drawn is black)+ P(the first ball drawn is black)P(the second ball drawn is red)

= (2/6)(4/5) + (4/6)(2/5)

=(8/30)+ (8/30)

= 16/30

= 8/15

**Question 4:**

A couple has two children,

(i) Find the probability that both children are males if it is known that at least one of the children is male.

(ii) Find the probability that both children are females if it is known that the elder child is a female.

**Solution:**

A couple has two children,

Let, the boy be denoted by b & girl be denoted by g

So, S = {(b, b) ,(b, g),(g, b), (g, g)}

To find probability that both children are males, if known that at least one of children is male

Let E : Both children are males

F : At least one child is male

To find P(E|F)

E : Both children are males

E = {(b, b)}

P(E) = 1/4

F : At least one child is male

F = {(b, g), (g, b), (b, b)}

P(F) = 3/4

E ∩ F = {(b, b)}

P(E ∩ F ) = 1/4

P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹))

= (1/4)/(3/4)

= 1/3

∴ Required Probability is 𝟏/𝟑

(ii) Find the probability that both children are females if it is known that the elder child is a female.

S = {(b, b) ,(b, g),(g, b), (g, g)}

To find the probability that both children are females, if from that the elder child is a female.

Let E : both children are females

F : elder child is a female

To find P(E|F)

E : both children are females

E = {(g, g)}

P(E) = 1/4

F : elder child is a female

F = {(g, b), (g, g)}

P(F) = 2/4=1/2

Also, E ∩ F = {(g, g)}

So, P(E ∩ F) = 1/4

P(E|F) = (𝑃(𝐸 ∩ 𝐹))/(𝑃(𝐹))

= (1/4)/(1/2)

= ½

**Question 5:**

One card is drawn from a well-shuffled pack of 52 cards. What is the probability that a card will be

(i) a diamond

(ii) Not an ace

(iii) a black card

(iv) not a diamond

**Solution:**

**(i) the probability that a card is a diamond**

We know that there are 13 diamond cards in a deck. Therefore, the required probability is:

P( getting a diamond card) = 13/52 = ¼

**(ii) the probability that a card is not an ace**

We know that there are 4 ace cards in a deck.

Therefore, the required probability is:

P(not getting an ace card) = 1-( 4/52)

= 1- (1/13)

=(13-1)/13

= 12/13

**(iii) the probability that a card is a black card**

We know that there are26 black cards in a deck.

Therefore, the required probability is:

P(getting a black card) = 26/52 = ½

**(iv) the probability that a card is not a diamond**

We know that there are 13 diamond cards in a deck.

We know that the probability of getting a diamond card is 1/4

Therefore, the required probability is:

P( not getting a diamond card) = 1- (1/4)

= (4-1)/4

= 3/4

**Question 6:**

A pack of 50 tickets is numbered from 1 to 50 and is shuffled. Two tickets are drawn at random. Find the probability that (i) both the tickets drawn bear prime numbers (ii) Neither of the tickets drawn bear prime numbers.

**Solution:**

The total number of tickets = 50

Prime numbers from 1 to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47.

The total number of prime numbers between 1 and 50 is 15.

(i) Probability that both tickets are drawn bears prime numbers:

P(Both tickets bearing prime numbers ) = ^{15}C_{2} / ^{50} C_{2}

= 3/35

Hence, the probability that both tickets are drawn bear prime numbers is 3/35.

(ii) Probability that neither of the tickets drawn bears prime numbers:

P ( Neither of the tickets bearing prime numbers) = ^{35}C_{2} / ^{50}C_{2}

= 17/35.

Therefore, the probability that neither of the tickets drawn bears a prime number is 17/35.

**Question 7:**

20 cards are numbered from 1 to 20. If one card is drawn at random, what is the probability that the number on the card is:

- Prime number
- Odd Number
- A multiple of 5
- Not divisible by 3

**Solution:**

Let S be the sample space.

S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

(1) Probability that the card drawn is a prime number:

Let E_{1} be the event of getting a prime number.

E_{1} = {2, 3, 5, 7, 11, 13, 17, 19}

Hence, P(E_{1}) = 8/20 = 2/5.

(2) Probability that the card drawn is an odd number:

Let E_{2} be the event of getting an odd number.

E_{2} = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

Hence, P(E_{2}) = 10/20 = 1/2.

(3) Probability that the card drawn is a multiple of 5

Let E_{3} be the event of getting a multiple of 5

E_{3} = {5, 10, 15, 20}

Hence, P(E_{3}) = 4/20 = 1/5.

(4) Probability that the card drawn is not divisible by 3:

Let E_{4} be the event of getting a number that is not divisible by 3.

E_{4} = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}

Hence, P(E_{4}) = 14/20 = 7/10.

### Important Questions for Class 11 Maths Chapter 16 Probability – Practice Questions

Practice the problems given below:

- What is the sample space for throwing a die?
- What is the sample space for tossing a coin?
- What is the sample space for the simultaneous toss of a die and a coin?
- In a single throw of two dice, find the probability of obtaining a total of 8.
- A bag contains 9 red and 12 white balls. One ball is drawn at random. Find the probability that the ball is drawn is red.
- What is the probability that the ordinary year has 53 Sundays?
- A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of its being a spade or a king.
- A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card?
- A number is chosen from the number 1 to 100. Find the probability of its being divisible by 4 or 6.
- Two cards are drawn at random from a pack of 52 cards. What is the probability that both the drawn cards are aces?
- A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?
- A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of being either red or king.

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