Two persons A and B take turns in throwing a | vedclass

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Question

(b) The probability of throwing $9$ with two dice $ = \frac{4}{{36}} = \frac{1}{9}$

$	herefore $ The probability of not throwing $9$ with two dice

Vedclass · Accepted Answer

$\frac{8}{{17}}$

Vedclass · Answer

(b) The probability of throwing $9$ with two dice $ = \frac{4}{{36}} = \frac{1}{9}$

$	herefore $ The probability of not throwing $9$ with two dice $ = \frac{8}{9}$

If $A$ is to win he should throw $9$ in $1^{st}$ or $3^{rd}$ or $5^{th}$ attempt

If $B$ is to win, he should throw, $9$ in $2^{nd} , 4^{th}$ attempt

$	herefore $ $B$'s chances

$ = \left( {\frac{8}{9}} ight)\,.\,\frac{1}{9} + {\left( {\frac{8}{9}} ight)^3}.\,\frac{1}{9} + .... = \frac{{\frac{8}{9} 	imes \frac{1}{9}}}{{1 - {{\left( {\frac{8}{9}} ight)}^2}}} = \frac{8}{{17}}$.

Two persons $A$ and $B$ take turns in throwing a pair of dice. The first person to through $9$ from both dice will be avoided the prize. If $A$ throws first then the probability that $B$ wins the game is

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