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# Probability that A speaks truth is $\large \frac{4}{5}$ . A coin is tossed. A reports that a head appears. The probability that actually there was head is?

$\begin{array} ((A) \frac{4}{5} \quad & (B) \frac{1}{2} \quad & (C) \frac{1}{5} \quad & (D) \frac{2}{5} \end{array}$

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $\;P(E_i/A)$=$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$
Let $E_1$ be the event that it comes up w a head. P ($E_1$) = $\large\frac{1}{2}$
Let $E_2$ be the event that it comes up w a head. P ($E_2$) = $\large\frac{1}{2}$
Let A be the event that A reports that a head appears.
P (A speaks the truth and a head appears) = P (A|$E_1$) = $\large\frac{4}{5}$
P (A does not speak the truth and a head does not appears) =1 - P (A|$E_1$) =1 - $\large\frac{4}{5} = \frac{1}{5}$
The probability that there is a head actually is P ($E_1$|A).
We can use Bayes theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P($E_1$|A) = $\large\frac{\frac{4}{5}.\frac{1}{2}}{\frac{4}{5}.\frac{1}{2} + \frac{1}{5}.\frac{1}{2}} = \frac{4}{4+1} = \frac{4}{5}$
edited Jun 19, 2013