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Q.
Let a biased coin be tossed $5$ times. If the probability of getting $4$ heads is equal to the probability of getting $5$ heads, then the probability of getting atmost two heads is:
$P ( H )= x , P ( T )=1- x$
$P (4 H , 1 T )= P (5 H )$
${ }^{5} C _{1}( x )^{4}(1- x )^{1}={ }^{5} C _{5} x ^{5}$
$5(1- x )= x$
$6 x =5=0 x =\frac{5}{6}$
$P ($ atmost $2 H )$
$= P ( OH , 5 T )+ P (1 H , 4 T )+ P (2 H , 3 T )$
$={ }^{5} C _{0}\left(\frac{1}{6}\right)^{5}+{ }^{5} C _{1} \frac{5}{6} \cdot\left(\frac{1}{6}\right)^{4}+{ }^{5} C _{2}\left(\frac{5}{6}\right)^{3}\left(\frac{1}{6}\right)^{3}$
$=\frac{1}{6^{5}}(1+25+250)=\frac{276}{6^{5}}$
$ = \frac{46}{6^4}$