The random variable X has a Binomial distribu | vedclass

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(B) Given $X$ is a random variable with $B(n=20, p=0.4)$.So,$n=20$,$p=0.4 = \frac{2}{5}$,and $q = 1 - p = \frac{3}{5}$.We need to calculate $5 - 5 P(X

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$43 \left(\frac{3}{5}\right)^{19}$

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(B) Given $X$ is a random variable with $B(n=20, p=0.4)$.So,$n=20$,$p=0.4 = \frac{2}{5}$,and $q = 1 - p = \frac{3}{5}$.We need to calculate $5 - 5 P(X \geq 2)$.$P(X \geq 2) = 1 - (P(X=0) + P(X=1))$.Substituting this into the expression:$5 - 5(1 - (P(X=0) + P(X=1))) = 5 - 5 + 5(P(X=0) + P(X=1)) = 5(P(X=0) + P(X=1))$.Using the Binomial probability formula $P(X=k) = {}^{n}C_{k} p^k q^{n-k}$:$P(X=0) = {}^{20}C_{0} (\frac{2}{5})^0 (\frac{3}{5})^{20} = (\frac{3}{5})^{20}$.$P(X=1) = {}^{20}C_{1} (\frac{2}{5})^1 (\frac{3}{5})^{19} = 20 	imes \frac{2}{5} 	imes (\frac{3}{5})^{19} = 8 	imes (\frac{3}{5})^{19}$.Now,$5(P(X=0) + P(X=1)) = 5 [(\frac{3}{5})^{20} + 8(\frac{3}{5})^{19}]$$= 5 [(\frac{3}{5}) 	imes (\frac{3}{5})^{19} + 8(\frac{3}{5})^{19}]$$= 5 [(\frac{3}{5} + 8) 	imes (\frac{3}{5})^{19}]$$= 5 [(\frac{3+40}{5}) 	imes (\frac{3}{5})^{19}]$$= 5 [\frac{43}{5} 	imes (\frac{3}{5})^{19}] = 43 	imes (\frac{3}{5})^{19}$.

The random variable $X$ has a Binomial distribution $B(20, 0.4)$. Then $5 - 5 P(X \geq 2) =$

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