Question

Let a biased coin be tossed $5$times. If the probability of getting $4\text{heads}$ is equal to the probability of getting $5\text{heads}$, then the probability of getting atmost two heads is:

Answer 1

Let probability of getting head $=p$

$\text{So,}{}^5{C}_4{p}^4(1-p)={}^5{C}_5{p}^5$

$\Rightarrow p=5(1-p)\Rightarrow p=\frac{5}{6}$

Probability of getting atmost two heads $=$

${}^5{C}_0(1-p{)}^5+{}^5{C}_{1}p(1-p{)}^4+{}^5{C}_2{p}^2(1-p{)}^3$

$=\frac{1+25+250}{6^5}=\frac{276}{6^5}=\frac{46}{6^4}$