Question

A and B roll a die successively till one of them gets a six. What are the chances of A' s winning if he starts the game?

• A. $\frac{4}{11}$
• B. $\frac{6}{11}$
• C. $\frac{8}{11}$
  
• D. $\frac{9}{11}$

Answer 1

The correct option is B $\frac{6}{11}$

Consider winning of $A$

Let S and F be the following events.

S: obtaining a six by $A$

F: not obtaining a six by $B$

A can will the following way:

(i) He obtains a six in the first trial

(ii) A fails to obtain a six, $B$ falls to obtain and then $A$ obtains a six

(iii) $A$ fails, then $B$ fails, then $A$ fails, then $B$ fails to obtain and finally, $A$ obtains a six and so on.

Obtaining a six by $A$ and not obtaining a six by $B$ are independent events.

$P\left(\text{A wins}\right)=P\left(S\right)+P\left(FFS\right)+P\left(FFFFS\right)+...⟹=\frac{1}{6}+\frac{1}{6}+times{\left(\frac{5}{6}\right)}^2+\frac{1}{6}{\left(\frac{5}{6}\right)}^4+...⟹=\frac{1}{6}\times \left[\frac{1}{1-{\left(\frac{5}{6}\right)}^2}\right]$

$[\because a+ar+a{r}^2+...term=\frac{a}{1-r},|r|<1]$

$⟹P\left(\text{A wins}\right)=\frac{\frac{1}{6}}{1-\frac{25}{36}}⟹=\frac{1}{6}\times \frac{36}{11}⟹P\left(\text{A wins}\right)=\frac{6}{11}$