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Class 12 Maths MCQ Questions of Probability with Answers?

Answer 1

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Practice MCQ Question for Class 12 Maths chapter-wise 

1. P(A ∩ B) is equal to:

(a) P(A) . P(B|A)
(b) P(B) . P(A|B)
(c) Both A and B
(d) None of these

2. If P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4, what is the value of P (A ∩ B)?

(a) 0.32
(b) 0.25
(c) 0.1
(d) 0.5

3. If P (A) = 6/11, P (B) = 5/11 and P (A ∪ B) = 7/11, what is the value of P(B|A)?

(a) 1/3
(b) 2/3
(c) 1
(d) none of these

4. Find P(E|F), where E: no tail appears, F: no head appears, when two coins are tossed in the air.

(a) 0
(b) 1/2
(c) 1
(d) None of the above

5. If P(A ∩ B) = 70% and P(B) = 85%, then P(A/B) is equal to:

(a) 17/14
(b) 14/17
(c) 7/8
(d) 1/8

6. If P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6. Find P(A ∪ B).

(a) 0.46
(b) 0.86
(c) 0.76
(d) 0.54

7. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

(a) 3/7
(b) 7/3
(c)1/7
(d) 1/3

8. If E and F are independent events, then;

(a) P(E ∩ F) = P(E)/ P(F)
(b) P(E ∩ F) = P(E) + P(F)
(c)P(E ∩ F) = P(E) . P(F)
(d) None of the above

9. If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by:

(a) 1+ P(A′) P (B′)
(b) 1− P(A′) P (B′)
(c) 1− P(A′) + P (B′)
(d) 1− P(A′) – P (B′)

10. The probability of solving the specific problems independently by A and B are 1/2 and 1/3 respectively. If both try to solve the problem independently, find the probability that exactly one of them

(a) 1
(b) 1/2
(c) 1/3
(d) 1/4

11. P(E ∩ F) is equal to

(a) P(E) . P(F|E)
(b) P(F) . P(E|F)
(c) Both (a) and (b)
(d) None of these

12. If A and B are two independent events, then

(a) P(A∩B) = P(a) × P(b)
(b) P(AB) = 1 – P(A’) P(B’)
(c) P(AB) = 1 + P (A’) P(B’) P(A’)
(d) P (AB) = P(A)/P(B′)

13. If the odd in favour of an event are 4 to 7, find the probability of its no occurence.

(a) 9/11
(b) 7/11
(c) 4/11
(d) 3/11

14. The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(a) 0
(b) 1/3
(c) 1/12
(d) 1/36

15. If two events are independent, then

(a) they must be mutually exclusive
(b) the sum of their probabilities must be equal to 1
(c) (a) and (b) both are correct
(d) None of the above is correct

16. If A and B are events such that P (A/B) = P(B/A), then

(a) A ⊂ B but A ≠ B
(b) A = B
(c) A ∩ B = ø
(d) P (A) = P (B)

17. If P(a) = 0,4, P(b) = 0.8 and P(B|A) = 0.6 then P(A∪B) is equal to

(a) 0.24
(b) 0.3
(c) 0.48
(d) 0.96

18. In a binomial distribution, the sum of its mean and variance is 1.8. Find the probability of two successes, if the event was conducted 5 times.

(a) 0.2048
(b) 0.85
(c) 0.32
(d) 0.96

19. If A and B are independent events such that P (A) > 0, P (B) > 0, then which one is not true:

(a) A and B are mutually exclusive
(b) B and A are mutually exclusive
(c) Both
(d) None

20.The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

(a) 28/256
(b) 219/256
(c) 128/256
(d) 37/256

Answer 2

1. Answer: (a) P(A) .P(B|A)

Explanation: If A and B are dependent events, then the probability of both events occurring simultaneously is given by: P(A ∩ B) = P(B) . P(A|B).

2. Answer: (a) 0.32

Explanation: Given, P (A) = 0.8, P (B) = 0.5 and P (B|A) = 0.4

By conditional probability, we have;

P (B|A) = P(A ∩ B)/P(A)

⇒ P (A ∩ B) = P(B|A). P(A) = 0.4 x 0.8 = 0.32

3. Answer: (b) 2/3

Explanation: P(B|A) = P(A ∩ B)/P(A) …(i)

Also,P(A ∩ B) = P(A) + P(B) – P(A U B)

= 6/11 + 5/11 – 7/11

= 4/11

Now putting the value of P(A ∩ B) in eq.(i), we get;

P(B|A) = (4/11)/(6/11) 

= 4/6

= 2/3

4. Answer: (a) 0

Explanation: E: no tail appears

And F: no head appears

⇒ E = {HH} and F = {TT}

⇒ E ∩ F = ϕ

As we know, two coins were tossed;

P(E) = 1/4

P(F) = 1/4

P(E ∩ F) = 0/4 = 0

Thus, by conditional probability, we know that;

P(E|F) = P(E ∩ F)/P(F)

= 0/(1/4)

= 0

5. Answer: (b) 14/17

Explanation: By conditional probability, we know;

P(A|B) = P(A ∩ B)/P(B)

= (70/100) x (100/85)

= 14/17

6. Answer: (b) 0.86

Explanation: P(A) = 0.4, P(B) = 0.7 and P(B/A) = 0.6

By conditional probability, we know;

P(B|A) = P(A ∩ B)/P(A)

⇒ 0.6 × 0.4 = P (A ∩ B)

⇒ P(A ∩ B) = 0.24

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

= 0.4 + 0.7 – 0.24

= 0.86

7. Answer: (a) 3/7

Explanation: Let E and F denote the events that the first and second ball drawn is black, respectively.

We need to find P(E ∩ F) or P (EF).

P(E) is the probability of black ball first drawn.

P(E) = 10/15

Now, 9 black balls are left in the urn.

P(F|E) = 9/14

By multiplication rule;

P (E ∩ F) = P (E) P(F|E)

= 10/15 x 9/14

= 3/7

8. Answer: (c)P(E ∩ F) = P(E) . P(F)

Explanation: P(E ∩ F) = P(E) . P(F)

Two events E and F are said to be independent if;

P(F|E) = P (F) given P (E) ≠ 0

and P (E|F) = P (E) given P (F) ≠ 0

Now, by the multiplication rule of probability, we have

P(E ∩ F) = P(E) . P (F|E) … (1)

If E and F are independent, then; P(E ∩ F) = P(E) . P(F)

9. Answer: (b) 1− P(A′) P (B′)

Explanation: P(at least one of A and B) = P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

= P(A) + P(B) − P(A) P(B)

= P(A) + P(B) [1−P(A)]

= P(A) + P(B). P(A′)

= 1− P(A′) + P(B) P(A′)

= 1− P(A′) [1− P(B)]

= 1− P(A′) P (B′)

10. Answer: (b) 1/2

Explanation: P(A) = 1/2

P(B) = 1/3

Since, A and B are independent events, therefore;

⇒ P (A ∩ B) = P (A). P (B)

⇒ P (A ∩ B) = ½ × 1/3 = 1/6

P (A’) = 1 – P (A) = 1 – 1/2 = 1/2

P (B’) = 1 – P (B) = 1 – 1/3 = 2/3

Now the probability that exactly one of them solved the problem is either the problem is solved by A and not B or vice versa.

P (A).P (B’) + P (A’).P (B)

= 1/2 (2/3) + 1/2 (1/3)

= 1/3 + 1/6 = 3/6

⇒ P (A).P (B’) + P (A’).P (B) = 1/2

Answer 3

11. Answer: (c) Both (a) and (b)

Explanation: We know that, conditional probability of event E given that F has occurred is denoted by P(E|F) and is given by

P(E|F) = P(E∩F)/P(F), P(F) ≠ 0

From this result, we can write

P(E∩F) =P(F) . P(E|F)…(i)

Also, we know that

P(F|E)=P(F∩E)/P(E), P(E)≠0

Thus, P(E∩F)=P(E).P(F/F)…(ii)

12. Answer: (a) P(A∩B) = P(a) × P(b)

Explanation: If A and B are independent events. It implies-

P(A ∩ B) = P(A)P(B)

13. Answer: (b) 7/11

Explanation: The odds in favour of the event is 4:7. Favourable elements for event A is 4x . Unfavourable elements for event A is 7x. Therefore, the probability that an event does not occurs

\(=\frac{7x}{4x+7x}\)

= 7/11

14. Answer: (d) 1/36

Explanation: When two dice are rolled, the number of outcomes is 36. The only even prime number is 2. Let E be the event of getting an even prime number on each die.

∴E = {(2,2)}

P(E) = 1/36

15. Answer: (d) None of the above is correct

Explanation: Two events A and B are said to be independent if the fact that one event has occurred does not affect the probability that the other event will occur. If whether or not one event occurs does affect the probability that the other event will occur, then the two events are said to be dependent.

16. Answer: (d) P (A) = P (B)

Explanation: It is given that, P(A∣B) = P(B∣A)

P(A∩B)/P(B) = P(A∩B)/P(A)

P(A) = P(B)

17. Answer: (d) 0.96

Explanation: P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6

We know that,

P(B|A) × P(A) = P(B ∩ A)

⇒ 0.6 × 0.4 = P(B ∩ A)

⇒ P(B ∩ A) = 0.24

Now,

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

[Additive Law of Probability]

= 0.4 + 0.8 – 0.24

= 0.96

18. Answer: (a) 0.2048

Explanation: Here np + npq =1.8 and n=5

⇒5p(1+q) =1.8

⇒(1−q)(1+q) = 0.36

⇒1−q² = 0.36

⇒q² = 0.64

⇒q =0.8 (Taking + ve value only)

∴n=5, p=0.2, g=0.8

∴ Probability of 2 successes = P(2) =⁵C₂p²q³

= 10(0.2)²(0.8)³

= 0.2048

19. Answer: (a) A and B are mutually exclusive

Explanation: Let A and B be two independent events.

∴P(A∩B)= P(A). P(B)≠0

(∵P(A) > 0,P(B) >0)

20. Answer: (a) 28/256

Explanation: mean = np = 4 and variance = npq = 2

∴p =q =12 and n = 8

∴P(2success) = ⁸C₂(1/2)⁶(1/2)²  

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