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Question

(A) Let the probability of getting a head be $x$ and the probability of getting a tail be $(1-x)$.Tom starts the game,so he wins if he gets a head on

Vedclass · Accepted Answer

$\frac{144}{625}$

Vedclass · Answer

(A) Let the probability of getting a head be $x$ and the probability of getting a tail be $(1-x)$.Tom starts the game,so he wins if he gets a head on his turn (1st,3rd,5th,... toss).The probability of Tom winning is $x + (1-x)^2 x + (1-x)^4 x + \dots = \frac{5}{8}$.This is an infinite geometric series with first term $a = x$ and common ratio $r = (1-x)^2$.The sum is $\frac{x}{1-(1-x)^2} = \frac{5}{8}$.$\frac{x}{1-(1-2x+x^2)} = \frac{5}{8} \Rightarrow \frac{x}{2x-x^2} = \frac{5}{8}$.Since $x 
eq 0$,we have $\frac{1}{2-x} = \frac{5}{8} \Rightarrow 8 = 10 - 5x \Rightarrow 5x = 2 \Rightarrow x = \frac{2}{5}$.Now,for $n=5$ tosses,the probability of getting exactly $r=3$ heads is given by the binomial distribution $P(X=r) = { }^n C_r p^r q^{n-r}$.Here $p = \frac{2}{5}$,$q = 1 - \frac{2}{5} = \frac{3}{5}$,$n=5$,$r=3$.$P(X=3) = { }^5 C_3 \left(\frac{2}{5}ight)^3 \left(\frac{3}{5}ight)^2 = 10 	imes \frac{8}{125} 	imes \frac{9}{25} = \frac{720}{3125} = \frac{144}{625}$.

Tom and Jerry play a game of alternately throwing an unfair coin. The first one to get a head wins. If Tom starts the game,he has a $62.5 \%$ chance of winning. Suppose this coin is tossed $5$ times,then the probability of getting exactly $3$ heads is

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