The probability of getting 4 heads in 8 throw | vedclass

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(D) The probability of getting $k$ successes in $n$ independent Bernoulli trials is given by the binomial distribution formula: $P(X = k) = {}^nC_k \c

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The probability of getting $k$ successes in $n$ independent Bernoulli trials is given by the binomial distribution formula: $P(X = k) = {}^nC_k \cdot p^k \cdot q^{n-k}$.Here,$n = 8$ (total number of throws),$k = 4$ (number of heads required),$p = \frac{1}{2}$ (probability of getting a head),and $q = 1 - p = \frac{1}{2}$ (probability of getting a tail).Substituting these values into the formula:$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^{8-4}$$P(X = 4) = {}^8C_4 \cdot (\frac{1}{2})^4 \cdot (\frac{1}{2})^4$$P(X = 4) = {}^8C_4 \cdot (\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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