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

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(D) Let $P(H) = x$ and $P(T) = 1 - x$.Given $P(4H, 1T) = P(5H)$.Using the binomial distribution formula,${}^{5}C_{4} x^{4}(1-x)^{1} = {}^{5}C_{5} x^{5

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

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(D) Let $P(H) = x$ and $P(T) = 1 - x$.Given $P(4H, 1T) = P(5H)$.Using the binomial distribution formula,${}^{5}C_{4} x^{4}(1-x)^{1} = {}^{5}C_{5} x^{5}$.$5(1-x) = x$.$5 - 5x = x \implies 6x = 5 \implies x = \frac{5}{6}$.Thus,$P(H) = \frac{5}{6}$ and $P(T) = \frac{1}{6}$.We need to find $P(	ext{at most } 2H) = P(0H) + P(1H) + P(2H)$.$P(0H) = {}^{5}C_{0} (\frac{5}{6})^{0} (\frac{1}{6})^{5} = \frac{1}{6^{5}}$.$P(1H) = {}^{5}C_{1} (\frac{5}{6})^{1} (\frac{1}{6})^{4} = 5 	imes \frac{5}{6^{5}} = \frac{25}{6^{5}}$.$P(2H) = {}^{5}C_{2} (\frac{5}{6})^{2} (\frac{1}{6})^{3} = 10 	imes \frac{25}{6^{5}} = \frac{250}{6^{5}}$.Summing these,$P(	ext{at most } 2H) = \frac{1 + 25 + 250}{6^{5}} = \frac{276}{6^{5}}$.Simplifying,$\frac{276}{6^{5}} = \frac{46 	imes 6}{6^{5}} = \frac{46}{6^{4}}$.

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 at most two heads is

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