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

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

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

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

- 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)

$\qquad=1-\large\frac{1}{9}$

$\qquad=\large\frac{8}{9}$

$P(F)=\large\frac{1}{9}$

P (\(\bar{F})\) = 1 - P(F)

$\qquad=1-\large\frac{1}{9}$

$\qquad=\large\frac{8}{9}$

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 $\large\frac{9}{17}$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...