Probability that A speaks truth is $\frac{4}{5}.$ A coin is tossed. A reports that a head appears. The probability that actually there was head is
A$\frac{4}{5}$
B$\frac{1}{2}$
C$\frac{1}{5}$
D$\frac{2}{5}$
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A$\frac{4}{5}$
a.$\frac{4}{5}$
Let A be the event that the man reports that head occurs in tossing a coin and let $E_1$ be the event that head occurs and $E_2$ be the event head does not occur.
$\text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_1)$ = P(A reports that head occurs when head had actually occur red on the coin) = $\frac{4}{5}$
$\text{P}(\text{A}|\text{E}_2)=$ P(A reports that head occurs when head had not occur red on the coin) $=1-\frac{4}{5}=\frac{1}{5}$
By Bayes’ theorem,
$ \text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+{\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)}}=\frac{\frac{1}{2}\times\frac{4}{5}}{\frac{1}{2}\times\frac{4}{5}+\frac{1}{2}\times\frac{1}{5}}=\frac{4}{4+1}=\frac{4}{5}$
Hence, option (A) is correct.
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