In tossing 10 coins,the probability of gettin | vedclass

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(B) The probability of getting exactly $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution formula: $P(X = k) = {}^{n}C_{k} \

Vedclass · Accepted Answer

$\frac{63}{256}$

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(B) The probability of getting exactly $k$ heads in $n$ tosses of a fair coin is given by the binomial distribution formula: $P(X = k) = {}^{n}C_{k} 	imes p^{k} 	imes q^{n-k}$.Here,$n = 10$,$k = 5$,$p = \frac{1}{2}$ (probability of head),and $q = 1 - p = \frac{1}{2}$ (probability of tail).Substituting these values:$P(X = 5) = {}^{10}C_{5} 	imes (\frac{1}{2})^{5} 	imes (\frac{1}{2})^{10-5}$$P(X = 5) = {}^{10}C_{5} 	imes (\frac{1}{2})^{10}$Calculating ${}^{10}C_{5} = \frac{10 	imes 9 	imes 8 	imes 7 	imes 6}{5 	imes 4 	imes 3 	imes 2 	imes 1} = 252$.Calculating $( \frac{1}{2} )^{10} = \frac{1}{1024}$.Thus,$P(X = 5) = 252 	imes \frac{1}{1024} = \frac{252}{1024} = \frac{63}{256}$.

In tossing $10$ coins,the probability of getting exactly $5$ heads is

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