Probability Ques 116
Assertion and Reason
For the following questions, choose the correct answer from the codes (a), (b), (c) and (d) defined as follows.
(a) Statement I is true, Statement II is also true; Statement II is the correct explanation of Statement I
(b) Statement I is true, Statement II is also true; Statement II is not the correct explanation of Statement I
(c) Statement I is true; Statement II is false
(d) Statement I is false; Statement II is true
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 $P\left(H _i / E\right)>P\left(E / H _i\right) \cdot P\left(H _i\right)$ for $\quad i=1,2, \ldots, n$
Statement II $\sum _{i=1}^{n} P\left(H _i\right)=1$
(2007, 3M)
Show Answer
Answer:
Correct Answer: 116.(d)
Solution:
Formula:
- Statement I If $P\left(H _i \cap E\right)=0$ for some $i$, then
$ P (\frac{H _i}{E})=P (\frac{E}{H _i})=0 $
If $P\left(H _i \cap E\right) \neq 0, \forall \quad i=1,2, \ldots, n$, then
$ \begin{aligned} & P (\frac{H _i}{E})=\frac{P\left(H _i \cap E\right)}{P\left(H _i\right)} \times \frac{P\left(H _i\right)}{P(E)} \\ & =\frac{P (\frac{E}{H _i}) \times P\left(H _i\right)}{P(E)}>P (\frac{E}{H _i}) \cdot P\left(H _i\right) \quad[\because 0<P(E)<1] \end{aligned} $
Hence, Statement I may not always be true.
Statement II Clearly, $H _1 \cup H _2 \cup \ldots \cup H _n=S$
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$ \Rightarrow \quad P\left(H _1\right)+P\left(H _2\right)+\ldots+P\left(H _n\right)=1 $
Hence, Statement II is ture.
BETA
