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 $n=$

Question

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

MARKS App · Accepted Answer

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 $\begin{aligned} P\left(E_1 \mid R\right) & =\frac{P\left(E_1\right) P\left(R \mid E_1\right)}{P\left(E_1\right) P\left(R \mid E_1\right)+P\left(E_2\right) P\left(R \mid E_2\right)} \ \Rightarrow \quad \frac{32}{33} & =\frac{2^n}{1+2^n} \Rightarrow n=5 \end{aligned}$

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