Question

The probability that $\mathrm{A}$speaks the truth is $\frac{4}{5}$. A coin is tossed. A report that a head appears. The probability that actually there was a head $?$

Answer 1

Find the required probability:

Step 1. Determine the probability that the head actually appears, if $A$ reports that a head appears.

Let ${\mathrm{E}}_1,{\mathrm{E}}_2$ and $\mathrm{A}$ be the events defined as follows: ${\mathrm{E}}_1=$ head occur, ${\mathrm{E}}_2=$ head does not occur, and $\mathrm{A}=$ person speaks that it is a head.

${\mathrm{E}}_1,{\mathrm{E}}_2$ are mutually exclusive and exhaustive.

$$\begin{gathered} \mathrm{P}\left({\mathrm{E}}_1\right)=\frac{1}{2} \\ \mathrm{P}\left({\mathrm{E}}_2\right)=1-\frac{1}{2}=\frac{1}{2} \end{gathered}$$

Now $\mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)=$ probability that the man reports that there is a head-on the coin given that a head has occurred on the coin $=\frac{4}{5}$ (probability that the person speaks the truth).

and $\mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_2}\right)=$ probability that the man reports that there is a head-on the coin given that head has not occurred on the coin (probability that the man does not speak truth $=1-\frac{4}{5}=\frac{1}{5}$.

According to Bayes' Theorem, the conditional probability of an event dependent on the occurrence of another event is equal to the likelihood of the second event multiplied by the probability of the first event.

Bayes' Theorem formula: $\mathrm{P}\left(\frac{{\mathrm{E}}_1}{\mathrm{A}}\right)=\frac{\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}\left({\displaystyle \frac{\mathrm{A}}{{\mathrm{E}}_1}}\right)}{\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_1}\right)+\mathrm{P}\left({\mathrm{E}}_2\right)\mathrm{P}\left(\frac{\mathrm{A}}{{\mathrm{E}}_2}\right)}$

$\begin{array}{rcl}\mathrm{P}\left({\mathrm{E}}_1/\mathrm{A}\right)& =& \frac{\frac{1}{2}\times \frac{4}{5}}{\frac{1}{2}\times \frac{4}{5}+\frac{1}{2}\times \frac{1}{5}}\\ & =& \frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{1}{2}}}\\ & =& \frac{4}{5}\end{array}$

Hence, the probability that the head actually appears, if $A$ reports that a head appears is $\frac{4}{5}.$