Tom and Jerry play a game of alternately throwing an unfair coin. First one to get head wins. If Tom starts…

Tom and Jerry play a game of alternately throwing an unfair coin. First one to get head wins.

The coin is unfair, hence, let, the probability of getting a head when the coin is tossed is $x$ then the probability of getting a tail when the coin is tossed is $\left(1-x\right).$

If Tom starts the game, he has $62.5\%=\frac{62.5}{100}=\frac{5}{8}$ chance of winning the game.

He can win the game in first toss of the coin or second toss or third toss and so on means the cases are $H$ or $TTH$ or $TTTTH$ or $TTTTTTH$ and so on.

Thus, the probability of Tom winning the game is

$x+\left(1-x\right)\times \left(1-x\right)\times x+\left(1-x\right)\times \left(1-x\right)\times \left(1-x\right)\times \left(1-x\right)\times x+...=\frac{5}{8}$

$\Rightarrow x+{\left(1-x\right)}^{2}x+{\left(1-x\right)}^{4}x+{\left(1-x\right)}^{6}x+...=\frac{5}{8}$

Using the sum of infinite terms of a geometric progression i.e. $a+ar+a{r}^2+a{r}^3+...=\frac{a}{1-r},$

$\Rightarrow \frac{x}{1-{\left(1-x\right)}^2}=\frac{5}{8}$

$\Rightarrow 8x=5\left[1-{\left(1-x\right)}^2\right]$

$\Rightarrow 8x=5\left[1-1+2x-x^2\right]$

$\Rightarrow 8x=5\left[2x-x^2\right]$

$\Rightarrow 8x=10x-5x^2$

$\Rightarrow 5x^2-2x=0$

$\Rightarrow x=0$ or $x=\frac{2}{5}.$

But $x=0$ is not possible, hence the probability of getting a head is $\frac{2}{5}.$

Now if this coin is tossed $5$ times, then the probability of getting exactly $3$ heads can be obtained by using Binomial distribution i.e.

$P\left(X=r\right)={}^{n}C_r{p}^r{q}^{n-r},$ where $n=5, p=\frac{2}{5}, q=\frac{3}{5}, r=3.$

$P\left(X=3\right)={}^{5}C_3·{\left(\frac{2}{5}\right)}^3·{\left(\frac{3}{5}\right)}^2$

$\Rightarrow P\left(X=3\right)=\left(\frac{5!}{3!\times 2!}\right)·\left(\frac{8}{125}\right)·\left(\frac{9}{25}\right)$

$\Rightarrow P\left(X=3\right)=\left(\frac{5\times 4\times 3!}{3!\times 2\times 1}\right)·\left(\frac{8}{125}\right)·\left(\frac{9}{25}\right)$

$\Rightarrow P\left(X=3\right)=\left(10\right)·\left(\frac{8}{125}\right)·\left(\frac{9}{25}\right)=\frac{144}{625}.$