Class 12 Maths Chapter 13 Probability MCQs with solutions are provided here for students. The multiple-choice questions are prepared as per the latest exam pattern. Class 12 Maths chapter-wise MCQs are given by our BYJU’S subject experts to help students score good marks in the board exams. Students can practice these questions that are formulated as per the CBSE syllabus (2022-2023) and NCERT curriculum. Let us see some of the MCQs and solve them to prepare for the chapter probability.
Probability Class 12 MCQs with Solutions
Find MCQs for Class 12 Chapter 13 Probability with solutions here.
Download PDF – Chapter 13 Probability MCQs
Q.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
Answer: C. Both A and B
Explanation: By multiplication theorem of probability.
Q.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
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
Q.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. â…“
B. â…”
C. 1
D. None of the above
Answer: B. â…”
Explanation: By definition of conditional probability we know;
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 = â…”
Q.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. ½
C. 1
D. None of the above
Answer: A. 0
Explanation: Given,
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) = ¼
P(F) = ¼
P(E ∩ F) = 0/4 = 0
Thus, by conditional probability, we know that;
P(E|F) = P(E ∩ F)/P(F)
= 0/(¼)
= 0
Q.5: If P(A ∩ B) = 70% and P(B) = 85%, then P(A/B) is equal to:
A. 17/14
B. 14/17
C. â…ž
D. â…›
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
Q.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
Answer: B. 0.86
Explanation: Given,
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
Q.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. â…“
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
Q.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
Answer: C. P(E ∩ F) = P(E) . P(F)
Explanation: 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)
Q.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′)
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′)
Q.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 solves the problem.
A. 1
B. ½
C. â…“
D. ¼
Answer: B. ½
Explanation: P(A) = ½
P(B) = â…“
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)
= ½ (2/3) + ½ (1/3)
= 1/3 + 1/6 = 3/6
⇒ P (A).P (B’) + P (A’).P (B) = ½
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