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Question 10

Mathematics Probability Medium

Let \( A_{1}, A_{2}, A_{3} \) are three events in a sample space with the condition: \( A_{1}\cap A_{2}\ne\emptyset \). then always

(A) \( P(A_{1}\cap A_{2}\cap A_{3})=P(A_{1})P(A_{2}/A_{1})P(A_{3}/(A_{1}\cap A_{2})) \)
(B) \( P(A_{1}\cap A_{2}\cap A_{3})=P(A_{1})P(A_{3}/A_{2}) \)
(C) \( P(A_{1}\cap A_{2}\cap A_{3})=P(A_{1})P(A_{2}/A_{3}) \)
(D) \( P(A_{1}\cap A_{2}\cap A_{3})=P(A_{1})P(A_{1}/A_{2})P(A_{1}/(A_{2}\cap A_{3})) \)
View Dynamic Solution & Explanation ▼
Correct Solution: Option A

Step-by-step Solution:

By the general multiplication rule of probability (or the chain rule for intersections of multiple events), the probability of the intersection of three events \( A_1, A_2, A_3 \) is evaluated sequentially: \[ P(A_1 \cap A_2 \cap A_3) = P(A_1) \times P(A_2 | A_1) \times P(A_3 | A_1 \cap A_2) \] This standard decomposition holds true as long as \( P(A_1 \cap A_2) > 0 \), which is effectively guaranteed by the given condition \( A_1 \cap A_2 \ne \emptyset \) (assuming non-empty sets have non-zero probabilities in this context). Thus, option A correctly expands the joint probability using conditional probabilities.

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