The probability of getting 4 heads in 8 throw | vedclass

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Question

(D) The probability of getting $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution formula: $P(X = k) = {}^nC_k 	imes p^k

Vedclass · Accepted Answer

$\frac{^8C_4}{2^8}$

Vedclass · Answer

(D) The probability of getting $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution formula: $P(X = k) = {}^nC_k 	imes p^k 	imes q^{n-k}$.Here,$n = 8$,$k = 4$,$p = \frac{1}{2}$ (probability of head),and $q = \frac{1}{2}$ (probability of tail).Substituting these values,we get:$P(X = 4) = {}^8C_4 	imes (\frac{1}{2})^4 	imes (\frac{1}{2})^{8-4}$$P(X = 4) = {}^8C_4 	imes (\frac{1}{2})^8$$P(X = 4) = \frac{{}^8C_4}{2^8}$.Thus,the correct option is $D$.

The probability of getting $4$ heads in $8$ throws of a coin is:

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