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(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \fra

Vedclass · Accepted Answer

$\frac{105}{512}$

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(B) For a fair coin,the probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = 1 - p = \frac{1}{2}$.Since the coin is tossed $n = 10$ times,this follows a binomial distribution $B(n, p) = B(10, \frac{1}{2})$.The probability of getting exactly $k = 6$ heads is given by the formula $P(X = k) = {}^{n}C_k \cdot p^k \cdot q^{n-k}$.Substituting the values: $P(X = 6) = {}^{10}C_6 \cdot (\frac{1}{2})^6 \cdot (\frac{1}{2})^{10-6}$.$P(X = 6) = {}^{10}C_6 \cdot (\frac{1}{2})^{10}$.Calculating ${}^{10}C_6 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.$P(X = 6) = 210 	imes \frac{1}{1024} = \frac{210}{1024} = \frac{105}{512}$.

$A$ coin is tossed $10$ times. The probability of getting exactly six heads is

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