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Find $P(3 \le X \le 7)$, where $X$ is a Binomial variable with parameters $n = 9$, $p = 0.86$.
Find $1 - p$.
$1 - p = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~}$ $1 - p = 0.14$
Hint: For a Binomial distribution, $P(X = r) = {}^n \mathrm{C}_r \times p^r \times (1-p)^{n-r}$.
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c1}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c2}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c3}}\hspace{100px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c4R}}\hspace{100px}}~}$ $P(X = 3) = 84 \times 0.636056 \times 0.00000753 = 0.00040229$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c1}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c2}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c3}}\hspace{100px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c4R}}\hspace{100px}}~}$ $P(X = 4) = 126 \times 0.54700816 \times 0.00005378 = 0.00370685$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c1}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c2}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c3}}\hspace{100px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c4R}}\hspace{100px}}~}$ $P(X = 5) = 126 \times 0.47042702 \times 0.00038416 = 0.02277062$
$P(X = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c0}}\hspace{35px}}~}) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c1}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c2}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c3}}\hspace{100px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c4R}}\hspace{100px}}~}$ $P(X = 6) = 84 \times 0.40456724 \times 0.002744 = 0.09325113$
$P(X = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c0}}\hspace{35px}}~}) = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c1}}\hspace{100px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c2}}\hspace{100px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c3}}\hspace{100px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c4R}}\hspace{100px}}~}$ $P(X = 7) = 36 \times 0.34792782 \times 0.0196 = 0.24549787$
Therefore, Hint: Add together the individual probabilities.
$P(3 \le X \le 7) = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$ $P(3 \le X \le 7) = 0.36562876$