Let a biased coin be tossed 5 times. If the p | vedclass

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Question

$P ( H )= x , P ( T )=1- x$

$P (4 H , 1 T )= P (5 H )$

${ }^{5} C _{1}( x )^{4}(1- x )^{1}={ }^{5} C _{5} x ^{5}$

$5(1- x )= x$

$6 x =5=0

Vedclass · Accepted Answer

$\frac{46}{6^{4}}$

Vedclass · Answer

$P ( H )= x , P ( T )=1- x$

$P (4 H , 1 T )= P (5 H )$

${ }^{5} C _{1}( x )^{4}(1- x )^{1}={ }^{5} C _{5} x ^{5}$

$5(1- x )= x$

$6 x =5=0 \quad x =\frac{5}{6}$

$P ($ atmost $2 H )$

$= P ( OH , 5 T )+ P (1 H , 4 T )+ P (2 H , 3 T )$

$={ }^{5} C _{0}\left(\frac{1}{6}\right)^{5}+{ }^{5} C _{1} \frac{5}{6} \cdot\left(\frac{1}{6}\right)^{4}+{ }^{5} C _{2}\left(\frac{5}{6}\right)^{3}\left(\frac{1}{6}\right)^{3}$

$=\frac{1}{6^{5}}(1+25+250)=\frac{276}{6^{5}}$

$=\frac{46}{6^{4}}$

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

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