Answer
a) $0.9298$
b) $9.1 \times 10^{-9}$
c) $0.01116$
d) $0.00163$
Work Step by Step
$X$ is binomial random variable with the parameters: $n=10, \quad p=0.1$
Calculate:
(a)
$$
\begin{aligned} \mathbb{P}(X \leq 2) &=\mathbb{P}(X=0)+\mathbb{P}(X=1)+\mathbb{P}(X=2)=\\ &=\left(\begin{array}{c}{10} \\ {0}\end{array}\right) 0.1^{0} 0.9^{10}+\left(\begin{array}{c}{10} \\ {1}\end{array}\right) 0.1^{1} 0.9^{9}+\left(\begin{array}{c}{10} \\ {2}\end{array}\right) 0.1^{2} 0.9^{8}=\\ &=0.9298 \end{aligned}
$$
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(b)
$$ \mathbb{P}(X>8) =\mathbb{P}(X=9)+\mathbb{P}(X=10)=$$
$$=\left(\begin{array}{c}{10} \\ {9}\end{array}\right) 0.1^{9} 0.9^{1}+\left(\begin{array}{c}{10} \\ {10}\end{array}\right) 0.1^{10} 0.9^{0}=9.1 \times 10^{-9} $$
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(c)
$$ \mathbb{P}(X=4) =\left(\begin{array}{c}{10} \\ {4}\end{array}\right) 0.1^{4} 0.9^{6}=0.01116 $$
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(d)
$\begin{aligned} \mathbb{P}(5 \leq X \leq 7) &=\mathbb{P}(X=5)+\mathbb{P}(X=6)+\mathbb{P}(X=7)=\\ &=\left(\begin{array}{c}{10} \\ {5}\end{array}\right) 0.1^{5} 0.9^{5}+\left(\begin{array}{c}{10} \\ {6}\end{array}\right) 0.1^{6} 0.9^{4}+\left(\begin{array}{c}{10} \\ {7}\end{array}\right) 0.1^{7} 0.9^{3}=\\ &=0.00163 \end{aligned}$
