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Q. Let $D_1, D_2, D_3$ be three tetrahedral dice where $D_1, D_2$ are printed with digits $3,3,4,6$ and $D_3$ is printed with $1,1,2,3$ on their faces. On throwing a dice each face is equally likely to fall. Outcome of the dice is the number appearing on the face which is downwards after the thrown. If $A$ and $B$ tosses a biased coin $\left( P ( H )=\frac{1}{3}\right)$ alternately starting with $A$. If $A$ throws tail before $B$ throws head then A wins the toss, they continue toss indefinitely, until one of them wins the toss. Now the winner randomly select 2 dice and rolled once. Three events $A , B$ and $E$ are defined as
A: 'A' wins the toss;
$B :$ ' $B$ ' wins the toss
$E :$ Sum on the dice is divisible by 3 .
$P \left(\frac{ B }{ E }\right)$ is equal to

Probability - Part 2

Solution:

$P ( A \text { win }) = T \text { or HTT or HTHTT or } \ldots \ldots \ldots . . $
$ =\frac{2}{3}+\frac{1}{3}\left(\frac{2}{3}\right)^2+\left(\frac{1}{3} \cdot \frac{2}{3}\right)^2 \cdot \frac{2}{3} \ldots \ldots=\frac{2 / 3}{1-2 / 9}=\frac{6}{7}$
$P ( B \text { win })=1-\frac{6}{7}=\frac{1}{7} $
$P \left(\frac{ B }{ E }\right)=\frac{ P ( B ) \cdot P ( E / B )}{ P ( A ) \cdot P ( E / B )+ P ( B ) \cdot P ( E / B )}=\frac{\frac{1}{7} \times \frac{17}{48}}{\frac{6}{7} \times \frac{17}{48}+\frac{1}{7} \times \frac{17}{48}}=\frac{1}{7}$