8 coins are tossed simultaneously. The probab | vedclass

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(D) The probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.For $n = 8$ tri

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$\frac{37}{256}$

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(D) The probability of getting a head in a single toss is $p = \frac{1}{2}$ and the probability of getting a tail is $q = \frac{1}{2}$.For $n = 8$ trials,the probability of getting $r$ heads is given by the binomial distribution formula: $P(X = r) = {}^nC_r p^r q^{n-r} = {}^8C_r (\frac{1}{2})^r (\frac{1}{2})^{8-r} = {}^8C_r (\frac{1}{2})^8$.We need to find the probability of getting at least $6$ heads,which is $P(X \ge 6) = P(X = 6) + P(X = 7) + P(X = 8)$.$P(X = 6) = {}^8C_6 (\frac{1}{2})^8 = \frac{8 	imes 7}{2 	imes 1} 	imes \frac{1}{256} = \frac{28}{256}$.$P(X = 7) = {}^8C_7 (\frac{1}{2})^8 = 8 	imes \frac{1}{256} = \frac{8}{256}$.$P(X = 8) = {}^8C_8 (\frac{1}{2})^8 = 1 	imes \frac{1}{256} = \frac{1}{256}$.Adding these probabilities: $P(X \ge 6) = \frac{28 + 8 + 1}{256} = \frac{37}{256}$.

$8$ coins are tossed simultaneously. The probability of getting at least $6$ heads is

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