Question

Class 11 Maths MCQ Questions of Probability with Answers?

Answer 1

Solve the under-given Class 11 Maths MCQ Questions of Probability with Answers and check your answers with the point-by-point answer here. Probability has a tremendous degree later on for higher assessments. If this part needs to comprehend by students, they can without a very remarkable stretch solve the powerful issues, taking into account this idea. 

Practicing the MCQ question for Class 11 Maths of Probability with answers will uphold your certainty accordingly helping you with scoring extraordinary in the exam. It is prescribed that students solve these MCQ Questions first and a short time later checks with the proper answer. This practice will help to acquire problem-solving capacities and build their confidence.

Practice MCQ Questions for class 11 Maths Chapter-Wise

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

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

3. What is the total number of sample spaces when a die is thrown 2 times?

(a) 6
(b) 12
(c) 18
(d) 36

4. What is the total number of sample spaces when a coin is tossed and a die is thrown?

(a) 6
(b) 12
(c) 8
(c) 16

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

(a) 1/6

(b) 1/36

(c) 1/18

(d) 3/28

6. 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

7. 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

8. 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

9. 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

10 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

11. 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

12. 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

13. 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

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

15. 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 2

1. Answer: (b) 143/11760

Explanation: 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 ×(¹¹C₂ / ⁴⁹C₂)
= 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

2. Answer: (c) 16/625

Explanation: 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

3. Answer: (d) 36

Explanation: The possible outcomes when a die is thrown are 1, 2, 3, 4, 5, and 6.
Given, a die is thrown two times.
Then, the total number of sample spaces = (6 × 6)
= 36

4. Answer: (b) 12

Explanation: The possible outcomes when a coin is tossed are Head (H) or Tail (T).
The possible outcomes when a die is thrown are 1, 2, 3, 4, 5, and 6.
Then, total number of space = (2 × 6) = 12

5. Answer: (b) 1/36

Explanation: 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

6. Answer: (b) 1/6

Explanation: 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

7. Answer: (d) 7/9

Explanation: 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

8. Answer: (d) 21

Explanation: 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

9. Answer: (a) 4/5

Explanation: Total number of ways of choosing two numbers out of six = 6C2 = (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

10. Answer: (c) e^-1/2/8

Explanation: 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

11. Answer: (d) 0.0545

 Explanation: 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

12. Answer: (b) 1/9

Explanation: One person can select one house out of 3 = ³C₁ = 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

13. Answer: (d) 4/9

Explanation: Total number of shocks = 5 + 4 = 9
Two shocks are pulled.
Now, P(Both are same color) = (⁵C₂ + ⁴C₂)/⁹C₂
= {(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

14. Answer: (c) 1/2

Explanation: 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

15. Answer: (b) 2/7

Explanation: 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

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