Probability 3 Question 29
29. If two events $A$ and $B$ are such that $P\left(A^{c}\right)=0.3, P(B)=0.4$ and $P\left(A \cap B^{c}\right)=0.5$, then $P\left[B /\left(A \cup B^{c}\right)\right]=\ldots$. $(1994,2 M)$
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Solution:
- $P\left(A^{c}\right)=0.3$
[given]
$$ \begin{aligned} & \Rightarrow \quad P(A)=0.7 \\ & P(B)=0.4 \\ & \Rightarrow \quad P\left(B^{c}\right)=0.6 \text { and } P\left(A \cap B^{c}\right)=0.5 \\ & \text { Now, } P\left(A \cup B^{c}\right)=P(A)+P\left(B^{c}\right)-P\left(A \cap B^{c}\right) \\ & \therefore P\left[B /\left(A \cup B^{c}\right]=\frac{P{B \cap\left(A \cup B^{c}\right) }}{P\left(A \cup B^{c}\right)}\right. \\ & =0.7+0.6-0.5=0.8 \\ & =\frac{P{(B \cap A) \cup\left(B \cap B^{c}\right) }}{0.8}=\frac{P{(B \cap A) \cup \varphi}}{0.8}=\frac{P(B \cap A)}{0.8} \end{aligned} $$
$$ \begin{aligned} & =\frac{1}{0.8}\left[P(A)-P\left(A \cap B^{c}\right)\right] \\ & =\frac{0.7-0.5}{0.8}=\frac{0.2}{0.8}=\frac{1}{4} \end{aligned} $$
