Question:
Let \(A_1,A_2,A_3\) be events in a sample space with \(A_1\cap A_2\neq \phi\). Then always:
Let \(A_1,A_2,A_3\) be events in a sample space with \(A_1\cap A_2\neq \phi\). Then always:
Show Hint
For three events,
\[
P(A\cap B\cap C)
=
P(A)P(B|A)P(C|A\cap B).
\]
This formula is frequently used in conditional probability questions.
Updated On: Jun 11, 2026
- \(P(A_1\cap A_2\cap A_3)=P(A_1)P(A_1/A_2)P(A_1/(A_2\cap A_3))\)
- \(P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2/A_1)P(A_3/(A_1\cap A_2))\)
- \(P(A_1\cap A_2\cap A_3)=P(A_1)P(A_3/A_2)\)
- \(P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2/A_3)\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
The multiplication law of probability states:
\[
P(A\cap B)
=
P(A)P(B|A).
\]
For three events,
\[
P(A\cap B\cap C)
=
P(A)P(B|A)P(C|A\cap B).
\]
Step 1: Apply multiplication theorem.
Taking \[ A=A_1,\qquad B=A_2,\qquad C=A_3, \] we obtain \[ P(A_1\cap A_2\cap A_3) = P(A_1) P(A_2|A_1) P(A_3|A_1\cap A_2). \]
Step 2: Compare with options.
This expression exactly matches option (B). \[ \boxed{\text{Option (B)}} \]
Step 1: Apply multiplication theorem.
Taking \[ A=A_1,\qquad B=A_2,\qquad C=A_3, \] we obtain \[ P(A_1\cap A_2\cap A_3) = P(A_1) P(A_2|A_1) P(A_3|A_1\cap A_2). \]
Step 2: Compare with options.
This expression exactly matches option (B). \[ \boxed{\text{Option (B)}} \]
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