There are two dice \(A\) and \(B\). Die \(A\) has 4 red and 2 white faces and \(B\) has 2 red and 4 white…
There are two dice \(A\) and \(B\). Die \(A\) has 4 red and 2 white faces and \(B\) has 2 red and 4 white faces. A coin is tossed once, if it shows head, die \(A\) is rolled, if it shows tail, die \(B\) is rolled, if the probability that die \(A\) is used is \(\left(\frac{32}{33}\right)\) when it is given that red turns up every time in first \(n\) throws, then \(n=\)
5
6
4
3
Solution
Let events, \(E_1\) : die \(A\) is used when head is appeared
\(E_2:\) die \(B\) is used when tail is appeared
\(R\) : red face appears.
\(\begin{array}{llll}
\because & \left(P\left(E_1 \mid R\right)\right. =\frac{32}{33} \\
\therefore & P\left(E_1\right) =\frac{1}{2}, P\left(E_2\right)=\frac{1}{2}, \\
& P\left(R \mid E_1\right) =\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \ldots n \text { times }=\left(\frac{2}{3}\right)^n
\end{array}\)
Similarly, \(P\left(R \mid E_2\right)=\left(\frac{1}{3}\right)^n\)
\(\because\) By Baye's theorem
\(\begin{aligned}
P\left(E_1 \mid R\right) & =\frac{P\left(E_1\right) P\left(R \mid E_1\right)}{P\left(E_1\right) P\left(R \mid E_1\right)+P\left(E_2\right) P\left(R \mid E_2\right)} \\
\Rightarrow \quad \frac{32}{33} & =\frac{2^n}{1+2^n} \Rightarrow n=5
\end{aligned}\)