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:

Let probability of getting head $=p$

\(\text { So, }{ }^{5} C_{4}~ p^{4}~(1-p)={ }^{5} C_{5}~ p^{5}\)

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

Probability of getting atmost two heads \(=\)

\({ }^{5} C_{0}~(1-p)^{5}+{ }^{5} C_{1}~ p~(1-p)^{4}+{ }^{5} C_{2}~ p^{2}~(1-p)^{3} \)

\(=\dfrac{1+25+250}{6^{5}}=\dfrac{276}{6^{5}}=\dfrac{46}{6^{4}}\\\)