Q34 / 66 · JEE Advanced 2007 (Paper 1) · Mathematics

Let $H_1, H_2, \ldots, H_n$ be mutually exclusive events with $P\left(H_i\right)>0, i=1,2, \ldots, n$. Let $E$ be any other event with $0 < P(E) < 1$. Statement I $P\left(H_i / E\right)>P\left(E / H_i\right) P\left(H_i\right)$ for $i=1,2, \ldots, n$. Statement II $\sum_{i=1}^n P\left(H_i\right)=1$

  • A
    Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
  • B
    Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement II
  • C
    Statement I is true, Statement II is false
  • Correct answer
    D
    Statement I is false, Statement II is true

Solution

Statement : I If $P\left(H_i \cap E\right)=0$ for some $i$, then $$ P\left(\frac{H_i}{E}\right)=P\left(\frac{E}{H_i}\right)=0 $$ If $P\left(H_i \cap E\right) \neq 0$ for $\forall i=1,2, \ldots . n$, then $$ \begin{aligned} P\left(\frac{H_i}{E}\right) & =\frac{P\left(H_i \cap E\right)}{P\left(H_i\right)} \times \frac{P\left(H_i\right)}{P(E)} \\ & \left.=\frac{P\left(\frac{E}{H_i}\right) \times P\left(H_i\right)}{P(E)}>P\left(\frac{E}{H_i}\right) \cdot P\left(H_i\right) \quad \text { [as } 0 < P(E) < 1\right] \end{aligned} $$ Hence, Statement I may not always be true. Statement II Clearly, $H_1 \cup H_2 \cup \ldots . \cup H_n=S$ (sample space) $$ \Rightarrow \quad P\left(H_1\right)+P\left(H_2\right)+\ldots+P\left(H_n\right)=1 . $$ Hence, Statement II is true.

Hence, the correct answer is option: 'D'.

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