Probability Ques 68
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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Answer:
Correct Answer: 68.$(\frac{1}{4})$
Solution:
Formula:
Set theoretical notation of probability and some important results:
- $P\left(A^{c}\right)=0.3$
[given]
$\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 $
$=\frac{P\{(B \cap A) \cup\left(B \cap B^{c}\right) \}}{0.8}=\frac{P\{(B \cap A) \cup \phi\}}{0.8}=\frac{P(B \cap A)}{0.8}$
$=\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}$
BETA
