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  • 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:<div class='qna-op

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:

Option: 1

\frac{275}{6^{5}}


Option: 2

\frac{36}{5^{4}}


Option: 3

\frac{181}{5^{5}}


Option: 4

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


Answers (1)

best_answer

Let \mathrm{P(H)=x \quad P(T)=1-x} \\

\mathrm{P(4 H, 1 T)=P(5 H)} \\

\mathrm{{ }^{5} C_{4} x^{4}(1-x)^1={ }^{5} C_{5} x^{5}}

\mathrm{5(1-x)=x} \\

\mathrm{6 {x}=5=0 \quad: x=\frac{5}{6}} \\

\mathrm{P(\text {atmost 2H)} }

\mathrm{=P(0 H, 5 T)+P(1 H, 4 T)+P(2 H, 3 T)} \\

\mathrm{={ }^{5} C_{0}\left(\frac{1}{6}\right)^{5}+{ }^{5} C_{1} \frac{5}{6}\left(\frac{1}{6}\right)^{4}+^{5} C_{2}\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)^{3}} \\

\mathrm{=\frac{1}{6^{5}} (1+25+250)=\frac{276}{6^5} }\\

\mathrm{= \frac{46}{6^{4}}}

Posted by

HARSH KANKARIA

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