Check the below NCERT MCQ Questions for Class 11 Maths Chapter 16 Probability with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Probability Class 11 Maths MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-11-maths-chapter-16/

## Probability Class 11 MCQs Questions with Answers

**MCQ On Probability Class 11 Question 1.**

Two cards from a pack of 52 cards are lost. One card is drawn from the remaining cards. If drawn card is diamond then the probability that the lost cards were both hearts is

(a) 143/1176

(b) 143/11760

(c) 143/11706

(d) 134/11760

## Answer

Answer: (b) 143/11760

Hint:

Total number of cards = 52

Two cards are lost.

So remaining cards = 50

Now one card is drawn.

Probability that it is a diamond card = 13/50

Now probability that both lost cards are heart = 13/50 ×(^{11}C_{2} / ^{49}C_{2})

= 13/50 ×[{(11×10)/2}/{(49×48/2)}]

= 13/50 ×{(11×10)/(49×48)}

= {(13×11×10)/(50×49×48)}

= {(13×11)/(5×49×48)}

= 143/11760

So probability that both lost card are heart = 143/11760

**Probability Class 11 MCQ Question 2.**

If four whole numbers taken at random are multiplied together, then the chance that the last digit in the product is 1, 3, 5, 7 is

(a) 16/25

(b) 16/125

(c) 16/625

(d) none of these

## Answer

Answer: (c) 16/625

Hint:

The last digit of the four whole number can be

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

The chance that any of the four numbers is divisible by 2 or 5 = 6/10 = 3/5

Hence, the chance that any of the four numbers is not divisible by 2 or 5 = 1 – 3/5 = 2/5

So, the chance that all of the four numbers are divisible by 2 or 5 = (2/5)×(2/5)×(2/5)×(2/5)

= 16/625

This is the chance that the last digit in the product will not be 0, 2, 4, 5, 6, 8 and this is also the chance that the last digit in the product is 1, 3, 7 or 9

**Probability MCQ Class 11 Question 3.**

Three identical dice are rolled. The probability that the same number will appear on each of them is

(a) 1/6

(b) 1/36

(c) 1/18

(d) 3/28

## Answer

Answer: (b) 1/36

Hint:

Total number of cases = 6³ = 216

The same number can appear on each of the dice in the following ways:

(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)

So, favourable number of cases = 6

Hence, required probability = 6/216 = 1/36

**Probability Class 11 Extra Questions Question 4.**

There are four machines and it is known that exactly two of them are faulty. They are tested, one by one, in a random order till both the faulty machines are identified. Then the probability that only two tests are needed is

(a) 1/3

(b) 1/6

(c) 1/2

(d) 1/4

## Answer

Answer: (b) 1/6

Hint:

First, we choose 1 machine out of given 4.

The probability that it is fault = 2/4 = 1/2

Now, we have to pick the second fault machine.

The probability that it is fault = 1/3

So, required probability = (1/2)×(1/3) = 1/6

**Class 11 Probability MCQ Question 5.**

Two unbiased dice are thrown. The probability that neither a doublet nor a total of 10 will appear is

(a) 3/5

(b) 2/7

(c) 5/7

(d) 7/9

## Answer

Answer: (d) 7/9

Hint:

When two dice are throw, then Total outcome = 36

A doublet: {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Favourable outcome = 6

Sum is 10: {(4, 6), (5, 5), (6, 4)}

Favourable outcome = 3

Again, A doublet and sum is 10: (5, 5)

Favourable outcome = 1

Now, P(either dublet or a sum of 10 appears) = P(A dublet appear) + P(sum is 10) – P(A dublet appear and sum is 10)

⇒ P(either dublet or a sum of 10 appears) = 6/36 + 3/36 – 1/36

= (6 + 3 – 1)/36

= 8/36

= 2/9

So, P(neither dublet nor a sum of 10 appears) = 1 – 2/9 = 7/9

**Probability Class 11 MCQ Questions Question 6.**

Two dice are thrown the events A, B, C are as follows A: Getting an odd number on the first die. B: Getting a total of 7 on the two dice. C: Getting a total of greater than or equal to 8 on the two dice. Then AUB is equal to

(a) 15

(b) 17

(c) 19

(d) 21

## Answer

Answer: (d) 21

Hint:

When two dice are thrown, then total outcome = 6×6 = 36

A: Getting an odd number on the first die.

A = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}

Total outcome = 18

B: Getting a total of 7 on the two dice.

B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

Total outcome = 6

C: Getting a total of greater than or equal to 8 on the two dice.

C = {(2, 6), (3, 5), (3, 6), (4, 4),(4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6),(6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

Total outcome = 15

Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ n(A ∪ B) = 18 + 6 – 3

⇒ n(A ∪ B) = 21

**Class 11 Maths Probability Extra Questions Question 7.**

Two numbers are chosen from {1, 2, 3, 4, 5, 6} one after another without replacement. Find the probability that the smaller of the two is less than 4.

(a) 4/5

(b) 1/15

(c) 1/5

(d) 14/15

## Answer

Answer: (a) 4/5

Hint:

Total number of ways of choosing two numbers out of six = ^{6}C_{2} = (6×5)/2 = 3×5 = 15

If smaller number is chosen as 3 then greater has choice are 4, 5, 6

So, total choices = 3

If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6

So, total choices = 4

If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6

So, total choices = 5

Total favourable case = 3 + 4 + 5 = 12

Now, required probability = 12/15 = 4/5

**Probability Class 11 Extra Questions With Solutions Question 8.**

The probability that when a hand of 7 cards is drawn from a well-shuffled deck of 52 cards, it contains 3 Kings is

(a) 1/221

(b) 5/716

(c) 9/1547

(d) None of these

## Answer

Answer: (c) 9/1547

Hint:

Total number of cards = 52

Number of king card = 4

Now, 7 cards are drawn from 52 cards.

P (3 cards are king) = {^{4}C_{3} × ^{48}C_{4}}/^{52}C_{7}

= {4×(48×47×46×45)/(4×3×2×1)}/{(52×51×50×49×48×47×46)/(7×6×5×4×3×2×1)}

= {4×(48×47×46×45)×(7×6×5×4×3×2×1)}/{(4×3×2×1)×{(52×51×50×49×48×47×46)}

= (7×6×5×4×45)/(52×51×50×49)

= (6×5×4×45)/(52×51×50×7)

= (6×4×45)/(7×52×51×10)

= (6×45)/(7×13×51×10)

= (6×3)/(7×13×17×2)

= (3×3)/(7×13×17)

= 9/1547

**MCQ Of Probability Class 11 Question 9.**

A certain company sells tractors which fail at a rate of 1 out of 1000. If 500 tractors are purchased from this company, what is the probability of 2 of them failing within first year

(a) e^{-1/2}/2

(b) e^{-1/2}/4

(c) e^{-1/2}/8

(d) none of these

## Answer

Answer: (c) e^{-1/2}/8

Hint:

This question is based on Poisson distribution.

Now, λ = np = 500×(1/1000) = 500/1000 = 1/2

Now, P(x = 2) = {e^{-1/2} × (1/2)²}/2! = e^{-1/2}/(4×2) = e^{-1/2}/8

**Probability Class 11 Questions Question 10.**

The probability that in a random arrangement of the letters of the word INSTITUTION the three T are together is

(a) 0.554

(b) 0.0554

(c) 0.545

(d) 0.0545

## Answer

Answer: (d) 0.0545

Hint:

Given word: INSTITUTION

Total letters = 11

The word contains 3 I, 2 N, 1 S, 3 T, 1 U and 1 O

Total number of arrangement = 11!/(3!×2!×3!) = 554400

Now, taken 3 T are together.

So total latter = 9

The number of favorable cases = 9!/(3!×2!) = 30240

Now, P(3 T are together) = 30240/554400 = 0.0545

**Class 11 Probability Questions Question 11.**

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is

(a) 2/9

(b) 1/9

(c) 8/9

(d) 7/9

## Answer

Answer: (b) 1/9

Hint:

One person can select one house out of 3 = ^{3}C_{1} = 3

So, three persons can select one house out of three = 3×3×3 = 27

Thus, probability that all the three can apply for the same house = 3/27 = 1/9

**Extra Questions Of Probability Class 11 Question 12.**

A bag contains 5 brown and 4 white socks . A man pulls out two socks. The probability that both the socks are of the same colour is

(a) 9/20

(b) 2/9

(c) 3/20

(d) 4/9

## Answer

Answer: (d) 4/9

Hint:

Total number of shocks = 5 + 4 = 9

Two shocks are pulled.

Now, P(Both are same color) = (^{5}C_{2} + ^{4}C_{2})/^{9}C_{2}

= {(5×4)/(2×1) + (4×3)/(2×1)}/{(9×8)/(2×1)}

= {(5×4) + (4×3)/}/{(9×8)

= (5 + 3)/(9×2)

= 8/18

= 4/9

**Probability Extra Questions Class 11 Question 13.**

When a coin is tossed 8 times getting a head is a success. Then the probability that at least 2 heads will occur is

(a) 247/265

(b) 73/256

(c) 247/256

(d) 27/256

## Answer

Answer: (c) 247/256

Hint:

Let x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)

The general form for probability of random variable x is

P(X = x) = ^{n}C_{x} × p^{x} × q^{n-x}

Now, in the question, we want at least two heads

Now, p = q = 1/2

So, P(X ≥ 2) = ^{8}C_{2} × (1/2)² × (1/2)^{8-2}

⇒ P(X ≥ 2) = ^{8}C_{2} × (1/2)² × (1/2)^{6}

⇒ 1 – P(X < 2) = ^{8}C_{0} × (1/2)^{0} × (1/2)^{8} + ^{8}C_{1} × (1/2)^{1} × (1/2)^{8-1}

⇒ 1 – P(X < 2) = (1/2)^{8} + 8 × (1/2)^{1} × (1/2)^{7}

⇒ 1 – P(X < 2) = 1/256 + 8 × (1/2)^{8}

⇒ 1 – P(X < 2) = 1/256 + 8/256

⇒ 1 – P(X < 2) = 9/256

⇒ P(X < 2) = 1 – 9/256

⇒ P(X < 2) = (256 – 9)/256

⇒ P(X < 2) = 247/256

**Probability Class 11 Important Questions Question 14.**

A couple has two children. The probability that both children are females if it is known that the elder child is a female is

(a) 0

(b) 1

(c) 1/2

(d) 1/3

## Answer

Answer: (c) 1/2

Hint:

Given, a couple has two children.

Let A denotes both children are females i.e. {FF}

Now, P(A) = (1/2)×(1/2) = 1/4

and B denotes elder children is a female i.e. {FF, FM}

P(B) = 1/4 + 1/4 = 1/2

Now, P(A ∩ B) = 1/4

Now, P(Both the children are female if elder child is female)

P(A/B) = P(A ∩ B)/P(B)

⇒ P(A/B) = (1/4)/(1/2)

⇒ P(A/B) = 1/2

**Important Questions Of Probability Class 11 Question 15.**

A certain company sells tractors which fail at a rate of 1 out of 1000. If 500 tractors are purchased from this company, what is the probability of 2 of them failing within first year

(a) e^{-1/2}/2

(b) e-^{-1/2}/4

(c) e^{-1/2}/8

(d) none of these

## Answer

Answer: (c) e^{-1/2}/8

Hint:

This question is based on Poisson distribution.

Now, λ = np = 500×(1/1000) = 500/1000 = 1/2

Now, P(x = 2) = {e^{-1/2} × (1/2)²}/2! = e^{-1/2} /(4×2) = e^{-1/2}/8

**Probability Important Questions Class 11 Question 16.**

A random variable X has poison distribution with mean 2. Then, P (X > 1.5) equals

(a) 1 – 3/e²

(b) 2/e²

(c) 3/e²

(d) 0

## Answer

Answer: (a) 1 – 3/e²

Hint:

Here m = 2

Now, P(X > 1.5) = ∑_{r} {(e^{-2} × 2^{r})/r!} {2 ≤ r ≤ ∞}

= e^{-2} {2²/2! + 2³/3! + 2^{4}/4! + …}

= e^{-2} {(1 + 2 /1! + 2²/2! + 2³/3! + …) – 1 – 2}

= e^{-2} (e² – 3)

= 1 – 3e^{-2}

= 1 – 3/e²

**Class 11 Probability Important Questions Question 17.**

Let A and B are two mutually exclusive events and if P(A) = 0.5 and P(B ̅) = 0.6 then P(A∪B) is

(a) 0

(b) 1

(c) 0.6

(d) 0.9

## Answer

Answer: (d) 0.9

Hint:

Given, A and B are two mutually exclusive events.

So, P(A ∩ B) = 0

Again given P(A) = 0.5 and P(B ̅) = 0.6

P(B) = 1 – P(B ̅) = 1 – 0.6 = 0.4

Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

⇒ P(A ∪ B) = P(A) + P(B)

⇒ P(A ∪ B) = 0.5 + 0.4 = 0.9

**Probability Questions Class 11 Question 18.**

The probability of getting 53 Sundays in a leap year is

(a) 1/7

(b) 2/7

(c) 3/7

(d) None of these

## Answer

Answer: (b) 2/7

Hint:

In a leap year, the total number of days = 366 days.

In 366 days, there are 52 weeks and 2 days.

Now two days may 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

Now there are total 7 possibilities, So total outcomes = 7

In 7 possibilities, Sunday came two times.

So, favorable case = 2

Hence, the probabilities of getting 53 Sundays in a leap year = 2/7

**Questions On Probability Class 11 Question 19.**

The probability of getting the number 6 at least once in a regular die if it can roll it 6 times?

(a) 1 – (5/6)^{6}

(b) 1 – (1/6)^{6}

(c) (5/6)^{6}

(d) (1/6)^{6}

## Answer

Answer: (a) 1 – (5/6)^{6}

Hint:

Let A is the event that 6 does not occur at all.

Now, the probability of at least one 6 occur = 1 – P(A)

= 1 – (5/6)^{6}

Question 20.

On his vacation, Rahul visits four cities (A, B, C, and D) in a random order. The probability that he visits A first and B last is

(a) 1/2

(b) 1/6

(c) 1/10

(d) 1/12

## Answer

Answer: (d) 1/12

Hint:

Total cities are 4 i.e. A, B, C, D

Given, Rahul visit four cities, So, n(S) = 4! = 24

Now, sample space IS:

S = {ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BDAC, BDCA, BCAD, BCDA, CABD, CADB, CBDA, CDAD, CDAB,CDBA, DABC, DACB, DBCA, DBAC, DCAB, DCBA}

Let G = Rahul visits A firsta and B last

⇒ G = {ACDB, ADCB}

⇒ n(G) = 2

So, P(G) = n(G)/n(S) = 2/24 = 1/12

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