For Binomial distribution B left(10, frac{1}{ | vedclass

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Question

(A) For a Binomial distribution $B(n, p)$,the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.Here,$n = 10$ and $p = \

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(A) For a Binomial distribution $B(n, p)$,the probability mass function is given by $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.Here,$n = 10$ and $p = \frac{1}{2}$,so $1-p = \frac{1}{2}$.Thus,$P(X = k) = \binom{10}{k} \left( \frac{1}{2} ight)^k \left( \frac{1}{2} ight)^{10-k} = \binom{10}{k} \left( \frac{1}{2} ight)^{10}$.We need to find $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$.$P(X=0) = \binom{10}{0} \left( \frac{1}{2} ight)^{10} = 1 	imes \left( \frac{1}{2} ight)^{10}$.$P(X=1) = \binom{10}{1} \left( \frac{1}{2} ight)^{10} = 10 	imes \left( \frac{1}{2} ight)^{10}$.$P(X=2) = \binom{10}{2} \left( \frac{1}{2} ight)^{10} = \frac{10 	imes 9}{2} 	imes \left( \frac{1}{2} ight)^{10} = 45 	imes \left( \frac{1}{2} ight)^{10}$.Adding these probabilities: $P(X \leq 2) = (1 + 10 + 45) \left( \frac{1}{2} ight)^{10} = 56 \left( \frac{1}{2} ight)^{10}$.Comparing this with $m \left( \frac{1}{2} ight)^{10}$,we get $m = 56$.

For Binomial distribution $B \left(10, \frac{1}{2} \right)$,if $P(X \leq 2) = m \left( \frac{1}{2} \right)^{10}$,then $m =$ . . . . . . .

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