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Answer: (a) \(2(0.8)^9\approx0.2684\) (b) \(1-(0.8)^{10}\approx0.8926\)

\(n=10\), \(p=0.2\), \(q=0.8\). \(P(X=x)={}^{10}C_x(0.2)^x(0.8)^{10-x}\), \(x=0,1,\dots,10\).

(a) \(P(X=1)={}^{10}C_1(0.2)(0.8)^9=2(0.8)^9\approx2\times0.1342=0.2684\).

(b) \(P(X\ge1)=1-P(X=0)=1-(0.8)^{10}\approx1-0.1074=0.8926\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report
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