Question

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

• A. $\frac{1}{7}$
• B. $\frac{6}{7}$
• C. $\frac{1}{56}$
• D. $\frac{1}{8}$

Answer 1

Answer: (A)

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