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Exercise 13.2 · Q8

Q.Let A and B be independent events with P(A)=0.3P(A) = 0.3 and P(B)=0.4P(B) = 0.4. Find

(i) P(A∩B)P(A \cap B)
(ii) P(A∪B)P(A \cup B)
(iii) P(A∣B)P(A|B)
(iv) P(B∣A)P(B|A)
Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★
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For independent events, the joint probability is the product of the individual probabilities. Using P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B), we get P(A∩B)=0.12P(A \cap B) = 0.12, P(A∪B)=0.58P(A \cup B) = 0.58, and both conditional probabilities equal the respective marginal probabilities: P(A∣B)=0.3P(A|B) = 0.3, P(B∣A)=0.4P(B|A) = 0.4.

The core idea here is event independence. When two events are independent, the occurrence of one has no effect on the probability of the other. This is the defining property, and it gives us a clean, direct way to compute everything.

Why independence matters:

If AA and BB are independent, then P(A∩B)=P(A)⋅P(B)P(A \cap B) = P(A) \cdot P(B). This is not a theorem we prove from something else — it is the definition of independence for two events. From this single fact, all other results follow naturally. The conditional probabilities become trivial: P(A∣B)=P(A)P(A|B) = P(A) and P(B∣A)=P(B)P(B|A) = P(B), because knowing BB happened gives no new information about AA, and vice versa.

Let’s work through each part step by step.

  1. Find P(A∩B)P(A \cap B) Since AA and BB are independent, the probability that both occur is simply the product of their individual probabilities.

P(A∩B)=P(A)⋅P(B)=0.3×0.4=0.12P(A \cap B) = P(A) \cdot P(B) = 0.3 \times 0.4 = 0.12

  1. Find P(A∪B)P(A \cup B) The union probability is given by the inclusion-exclusion principle:

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

We already have P(A∩B)=0.12P(A \cap B) = 0.12, so:

P(A∪B)=0.3+0.4−0.12=0.58P(A \cup B) = 0.3 + 0.4 - 0.12 = 0.58

  1. Find P(A∣B)P(A|B) The conditional probability of AA given BB is defined as:

P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

Substituting the values:

P(A∣B)=0.120.4=0.3P(A|B) = \frac{0.12}{0.4} = 0.3 …

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