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Find $P(1 \le X \le 5)$, where $X$ is a Binomial variable with parameters $n = 8$, $p = 0.11$.
Find $1 - p$.
$1 - p = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~}$ $1 - p = 0.89$
Hint: For a Binomial distribution, $P(X = r) = {}^n \mathrm{C}_r \times p^r \times (1-p)^{n-r}$.
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c1}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c2}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c3}}\hspace{100px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c4R}}\hspace{100px}}~}$ $P(X = 1) = 8 \times 0.11 \times 0.44231335 = 0.38923575$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c1}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c2}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c3}}\hspace{100px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c4R}}\hspace{100px}}~}$ $P(X = 2) = 28 \times 0.0121 \times 0.49698129 = 0.16837726$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c1}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c2}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c3}}\hspace{100px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c4R}}\hspace{100px}}~}$ $P(X = 3) = 56 \times 0.001331 \times 0.55840594 = 0.04162135$
$P(X = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c0}}\hspace{35px}}~}) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c1}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c2}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c3}}\hspace{100px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c4R}}\hspace{100px}}~}$ $P(X = 4) = 70 \times 0.00014641 \times 0.62742241 = 0.00643026$
$P(X = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c0}}\hspace{35px}}~}) = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c1}}\hspace{100px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c2}}\hspace{100px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c3}}\hspace{100px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c4R}}\hspace{100px}}~}$ $P(X = 5) = 56 \times 0.00001611 \times 0.704969 = 0.0006358$
$P(1 \le X \le 5) = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$ $P(1 \le X \le 5) = 0.60630042$