A box contains N coins of which m are fair and rest are biased. The probability of getting a head when a fair coin is tossed is $\frac{1}{2}$ , while it is $\frac{2}{3}$ when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. If the probability that the coin drawn is fair is f(N, m) ,then the value of 43 f(20,12) must be

Difficult

Solution

Let E₁ E₂ and A denote the following events
E₁: coin selected is fair
E₂: coin selected is biased
A:The first toss results in a head and the second toss results in a tail.
$\begin{array}{l} P (E_{1}) = \frac{m}{N}, P (E_{2}) = \frac{N - m}{N}, P (\frac{A}{E_{1}}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \\ and P (\frac{A}{E_{2}}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \end{array}$
By Baye's theorem
$\begin{array}{l} p (\frac{E_{1}}{A}) = \frac{P (E_{1}) \cdot P (\frac{A}{E_{1}})}{P (E_{1}) \cdot P (\frac{A}{E_{1}}) + P (E_{2}) \cdot P (\frac{A}{E_{2}})} \\ = \frac{\frac{m}{N} \times \frac{1}{4}}{\frac{m}{N} \times \frac{1}{4} + (\frac{N - m}{N}) \times \frac{2}{9}} \\ = \frac{9 m}{8 N + m} = f (N, m) \end{array}$
$f (20, 12) = \frac{9 \times 12}{8 \times 20 + 12} = \frac{108}{172} = \frac{27}{43}$
$\Rightarrow 43 f (20, 12) = 27$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME