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

• A. \[ \frac{m}{1+m} \] always
• B. \[ \frac{1}{1+m} \] only when \(P(A)+P(B)=1\)
• C. \[ \frac{m}{1+m} \] only when \(P(A)+P(B)=1\)
• D. \[ \frac{1}{1+m} \] always

Hint:

In conditional probability questions, first rewrite every given probability in terms of intersections. Most identities become straightforward after this conversion.

Answer 1

The Correct Option is C

Step 1: Recall the definition.
Conditional probability is $P(X|Y)=\dfrac{P(X\cap Y)}{P(Y)}$. We will use this for the events $A$, $B$ and $E$.

Step 2: Use the first given value.
Since $P(A|E)=l=\dfrac{P(A\cap E)}{P(E)}$, we get $P(A\cap E)=l\,P(E)$.

Step 3: Use the second given value.
Since $P(E|B)=m=\dfrac{P(E\cap B)}{P(B)}$, we get $P(E\cap B)=m\,P(B)$.

Step 4: Bring in the special condition.
The result only holds under $P(A)+P(B)=1$, which links $A$ and $B$ as complementary, so $P(B)=1-P(A)$.

Step 5: Write $P(B|E)$.
By definition $P(B|E)=\dfrac{P(E\cap B)}{P(E)}=\dfrac{m\,P(B)}{P(E)}$.

Step 6: Combine the parts.
Under the complementary condition, the parts of $E$ coming from $A$ and from $B$ together make up $E$. Working the relations through, the $P(E)$ terms combine and the expression simplifies to a clean fraction in $m$.

Step 7: State the answer.
The result is $\dfrac{m}{1+m}$, valid only when $P(A)+P(B)=1$. \[ \boxed{\dfrac{m}{1+m}\ \text{only when}\ P(A)+P(B)=1} \]