Probability that \mathrm{A} speaks truth is 4 / 5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
Probability that \mathrm{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. 4 / 5

B. 1 / 2

C. 1 / 5

D. 2 / 5

Solution:

A. 4 / 5

Explanation:

Let E_{1} represent the event in which A speaks the truth, E_{2} represent the event in which A lies, and X represent the event in which it appears head.

As a result, P\left(E{1}\right)=4 / 5.

E_{1} and E_{2} are events that are complementary to one another.

Then \mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)=1

\Rightarrow \mathrm{P}\left(\mathrm{E}_{2}\right)=1-\mathrm{P}\left(\mathrm{E}_{1}\right)

\Rightarrow \mathrm{P}\left(\mathrm{E}_{2}\right)=1-4 / 5=1 / 5

If you toss a coin, it may land on its head or tail.

As a result, whether A tells the truth or lies, the likelihood of getting head is 1 / 2.

P\left(X \mid E_{1}\right)=P\left(X \mid E_{2}\right)=1 / 2

Given that A tells the truth, the probability that there was head is P\left(E_{1} \mid X\right).

We have used Bayes’ theorem to arrive at our conclusion.

\mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{X}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\mathrm{X} \mid \mathrm{E}_{2}\right)}

Substituting the values, we now have

=\frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{4}{5} \cdot \frac{1}{2}+\frac{1}{5} \cdot \frac{1}{2}}

=\frac{\frac{2}{5}}{\frac{2}{5}+\frac{1}{10}}=\frac{\frac{2}{5}}{\frac{5}{10}}=\frac{4}{5}

\Rightarrow \mathrm{P}\left(\mathrm{E}_{1} \mid \mathrm{X}\right)=\frac{4}{5}

Consequently, the right response is (A).