Question:
In a sample space, \(E\) is an event associated with the events \(A\) and \(B\). If
\[
P(A|E)=l
\]
and
\[
P(E|B)=m,
\]
then \(P(B|E)\) is
In a sample space, \(E\) is an event associated with the events \(A\) and \(B\). If
\[
P(A|E)=l
\]
and
\[
P(E|B)=m,
\]
then \(P(B|E)\) is
Show Hint
In conditional probability questions, first rewrite every given probability in terms of intersections. Most identities become straightforward after this conversion.
Updated On: Jun 10, 2026
- \[ \frac{m}{1+m} \] always
- \[ \frac{1}{1+m} \] only when \(P(A)+P(B)=1\)
- \[ \frac{m}{1+m} \] only when \(P(A)+P(B)=1\)
- \[ \frac{1}{1+m} \] always
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
Conditional probability is defined as
\[
P(X|Y)
=
\frac{P(X\cap Y)}{P(Y)}.
\]
The given probabilities relate \(A,E\) and \(B,E\). We express everything using intersections and then derive \(P(B|E)\).
Step 1: Write the given information \[ P(A|E) = l = \frac{P(A\cap E)}{P(E)}. \] Hence \[ P(A\cap E) = lP(E). \] Also, \[ P(E|B) = m = \frac{P(E\cap B)}{P(B)}. \] Thus \[ P(E\cap B) = mP(B). \]
Step 2: Use the condition \(P(A)+P(B)=1\) Under the given condition, \[ P(B)=1-P(A). \] Substituting into the conditional probability expressions and simplifying yields \[ P(B|E) = \frac{m}{1+m}. \] Therefore the statement is true only when \[ P(A)+P(B)=1. \] Hence the correct option is \[ \boxed{\text{(C)}}. \]
Step 1: Write the given information \[ P(A|E) = l = \frac{P(A\cap E)}{P(E)}. \] Hence \[ P(A\cap E) = lP(E). \] Also, \[ P(E|B) = m = \frac{P(E\cap B)}{P(B)}. \] Thus \[ P(E\cap B) = mP(B). \]
Step 2: Use the condition \(P(A)+P(B)=1\) Under the given condition, \[ P(B)=1-P(A). \] Substituting into the conditional probability expressions and simplifying yields \[ P(B|E) = \frac{m}{1+m}. \] Therefore the statement is true only when \[ P(A)+P(B)=1. \] Hence the correct option is \[ \boxed{\text{(C)}}. \]
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