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A and B throw a pair of die turn by turn. The first to throw 9 is awarded a prize. If A starts the game, What is the probability of A getting the prize?

(A) $\large\frac{11}{17}$

(B) $\large\frac{4}{17}$

(C) $\large\frac{6}{17}$

(D) $\large\frac{9}{17}$

Can you answer this question?

1 Answer

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  • P (\(\bar{E})\) = 1 - P(E)
  • P (\(\bar{F})\) = 1 - P(F)
  • Sum of intinite G.P.$=\large\frac{a}{1-r}$ where $a=1^{st}\:term\:and\:r=common\:ratio.$
Step 1:
Let us define the following events as
$E$=The person A gets 9
$F$=The person B gets 9
Since a pair of die is thrown the sample space is 36.
Getting a '9' is $\{(5,4),(4,5),(6,3),(3,6)\}$
$\Rightarrow 4$
$\therefore P(E)=\large\frac{4}{36}=\frac{1}{9}$
P (\(\bar{E})\) = 1 - P(E)
P (\(\bar{F})\) = 1 - P(F)
Step 2:
A wins if he throws a 'nine' in 1st,or 3rd,5th or......throws
Probability of $A$ throwing a 'nine' in first throw = $\large\frac{1}{9}$
A will get third throw if he fails in first and B fails in the second throw.
$\therefore$Probability of $A$ winning in the third throw =$P( E\cap\bar F\cap \bar E)$
$\Rightarrow \large\frac{8}{9}\times\frac{8}{9}\times\frac{1}{9}$
$\Rightarrow (\large\frac{8}{9})^2\times\frac{1}{9}$
Step 3:
Probability of A winning in fifth row =$P(\bar E\cap \bar F\cap \bar E\cap\bar F\cap E)$
$\Rightarrow \big(\large\frac{8}{9})^4\times \frac{1}{9}$
and so on.
Step 4:
Hence the probability of A winning is
$P(E)+P(\bar E\cap \bar F\cap E)+P(\bar E\cap \bar F \cap\bar E\cap\bar F\cap E)$
$=\large\frac{1}{9}+(\large\frac{8}{9})^2.\frac{1}{9}+(\large\frac{8}{9})^4.\frac{1}{9}+....$ which is an infinite G.P.
We know that Sum of intinite G.P.$=\large\frac{a}{1-r}$ where $a=1^{st}\:term\:and\:r=common\:ratio.$
$\Rightarrow\:P(A$ winning)$ = \large\frac{1/9}{1-(8/9)^2}$
$= \large\frac{1/9}{17/81}$
$\Rightarrow \large\frac{9}{17}$
Hence the probability of A winning is $\large\frac{9}{17}$
answered Sep 25, 2013 by sreemathi.v
edited Mar 26, 2014 by rvidyagovindarajan_1

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