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NCERT Exemplar · Q41

Q.Let P(A)=713P(A) = \dfrac{7}{13}, P(B)=913P(B) = \dfrac{9}{13} and P(A∩B)=413P(A \cap B) = \dfrac{4}{13}. Then P(A′∣B)P(A' \mid B) is equal to
(A) 613\dfrac{6}{13}
(B) 413\dfrac{4}{13}
(C) 49\dfrac{4}{9}
(D) 59\dfrac{5}{9}

Uttarakhand UbseMCQ· 1mImportance★★★★★
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The key idea is to use the definition of conditional probability: P(A′∣B)=P(A′∩B)P(B)P(A' \mid B) = \frac{P(A' \cap B)}{P(B)}. Since A′∩BA' \cap B is just the part of BB that is not in AA, we have P(A′∩B)=P(B)−P(A∩B)P(A' \cap B) = P(B) - P(A \cap B). Substituting the given values gives 59\frac{5}{9}, which corresponds to option (D).

We are asked for P(A′∣B)P(A' \mid B) — the probability that AA does not happen, given that BB has happened. This is a classic conditional probability problem.

Why this approach works:

Conditional probability shrinks the sample space to the event we are conditioning on (here, BB). Within that restricted space, we want the portion where AA is false. That portion is exactly BB minus the part where AA and BB both occur — i.e., B∖(A∩B)B \setminus (A \cap B). So the numerator becomes P(B)−P(A∩B)P(B) - P(A \cap B).

Let’s work through it step by step.

  1. Recall the definition of conditional probability: For any two events XX and YY with P(Y)>0P(Y) > 0,

P(X∣Y)=P(X∩Y)P(Y).P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}.

Here, X=A′X = A' and Y=BY = B, so

P(A′∣B)=P(A′∩B)P(B).P(A' \mid B) = \frac{P(A' \cap B)}{P(B)}.

  1. Find P(A′∩B)P(A' \cap B): The event A′∩BA' \cap B means “BB occurs but AA does not.” This is exactly BB minus the overlap A∩BA \cap B. Since A∩BA \cap B is a subset of BB, we have

P(A′∩B)=P(B)−P(A∩B).P(A' \cap B) = P(B) - P(A \cap B).

  1. Substitute the given values: P(B)=913P(B) = \frac{9}{13} and P(A∩B)=413P(A \cap B) = \frac{4}{13}. So

P(A′∩B)=913−413=513.P(A' \cap B) = \frac{9}{13} - \frac{4}{13} = \frac{5}{13}.

  1. Now compute the conditional probability: …

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