Mathematics - Probability Question with Solution | TestHub

Given,

$P\left(H\right)=\frac{1}{3}$

So, $P\left(T\right)=1-\frac{1}{3}=\frac{2}{3}$

Now probability of the event is given by,

$P\left(E\right)=P\left(HH\right)+P\left(THH\right)+P\left(HTHH\right)+P\left(THTHH\right)+P\left(HTHTHH\right)+P\left(THTHTHH\right)+\dots$

$\Rightarrow P\left(E\right)=\frac{1}{3}\times \frac{1}{3}+\frac{2}{3}\times \frac{1}{3}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{3}\times \frac{1}{3}\times \frac{1}{3}+\frac{4}{3^5}+\frac{4}{3^6}+\frac{8}{3^7}+\frac{8}{3^8}+\dots$

$\Rightarrow P\left(E\right)=\frac{1}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}+\frac{4}{3^5}+\frac{4}{3^6}+\frac{8}{3^7}+\frac{8}{3^8}+\dots$

$\Rightarrow P\left(E\right)=\left(\frac{1}{3^2}+\frac{2}{3^4}+\frac{4}{3^6}+\dots .\right)+\left(\frac{2}{3^3}+\frac{4}{3^5}+\frac{8}{3^7}+\dots .\right)$

$\Rightarrow P\left(E\right)=\left(\frac{\frac{1}{3^2}}{\left(1-\frac{2}{3^2}\right)}\right)+\left(\frac{\frac{2}{3^3}}{\left(1-\frac{2}{3^2}\right)}\right)$

$\Rightarrow P\left(E\right)=\frac{1}{7}+\frac{2}{21}=\frac{5}{21}$