Probability that A speaks truth is \(\frac{4}{5}\). He tosses a coin and reports that a head appears. The probability that actually there was a head, is
Answer & explanation
Correct answer: option 1
Let E1 and E2 be the events such that
E1:A speaks truth
E2:A speaks false
Let X be the event that a head appears.
$P(E_1)=\frac{4}{5}$
$∴P(E_2)=1−P(E_1)=1−\frac{4}{5}=\frac{1}{5}$
If a coin is tossed, then it may result in either head (H) or tail (T).
only there will be two outcomes.
The probability of getting a head is $\frac{1}{2}$ whether A speaks truth or not.
$∴P(\frac{X}{E_1})=P(\frac{X}{E_2})=\frac{1}{2}$
The probability that there is actually a head is given by P(E1/X).
$∴P(E_1/X)=\frac{P(E_1).P(X/E_1)}{P(E_1).P(X/E_1)+P(E_2).P(X/E_2)}$
$=\frac{\frac{4}{5}×\frac{1}{2}}{\frac{4}{5}×\frac{1}{2}+\frac{1}{5}×\frac{1}{2}}=\frac{\frac{1}{2}×\frac{4}{5}}{\frac{1}{2}(\frac{4}{5}+\frac{1}{5})}⇒\frac{\frac{4}{5}}{\frac{5}{5}}=\frac{4}{5}$