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Probability that A speaks truth is 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) $\frac{2}{5}$
(a) $\frac{4}{5}$
(b) $\frac{1}{2}$
(c) $\frac{1}{5}$
(d) $\frac{2}{5}$
Solution:
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Verified Answer
Let $\mathrm{A}$ be the event that the man reports that head occurs in tossing a coin and let $\mathrm{E}_1$ be the event that head occurs and $E_2$ be the event head does not occurs.
$P\left(E_1\right)=\frac{1}{2}=P\left(E_2\right)$
$P\left(A \mid E_1\right)=$ Probability that A reports that head occurs when head has actually occurs red on the coin $=$ Probability that the man speaks truth $=\frac{4}{5}$ $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_2\right)=$ Probability that $\mathrm{A}$ reports that head occurs when head has not occurred on the coin $=$ Probability that the man does not speak the truth
$=1-\frac{4}{5}=\frac{1}{5}$
By Baye's theorem
$\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)=$ Probability that the report of A that head occurs, is actually a head
$$
=\frac{P\left(E_1\right) P\left(A / E_1\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}=\frac{4}{5}
$$
Thus option (a) is correct.
$P\left(E_1\right)=\frac{1}{2}=P\left(E_2\right)$
$P\left(A \mid E_1\right)=$ Probability that A reports that head occurs when head has actually occurs red on the coin $=$ Probability that the man speaks truth $=\frac{4}{5}$ $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_2\right)=$ Probability that $\mathrm{A}$ reports that head occurs when head has not occurred on the coin $=$ Probability that the man does not speak the truth
$=1-\frac{4}{5}=\frac{1}{5}$
By Baye's theorem
$\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)=$ Probability that the report of A that head occurs, is actually a head
$$
=\frac{P\left(E_1\right) P\left(A / E_1\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) P\left(A / E_2\right)}=\frac{4}{5}
$$
Thus option (a) is correct.
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