Formulaic Maths Problems
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Find $P(1 < X \le 5)$, where $X$ is a Binomial variable with parameters $n = 9$, $p = 0.91$.
Find $1 - p$.
$1 - p = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~}$
$1 - p = [0.09]$
Hint: For a Binomial distribution, $P(X = r) = {}^n \mathrm{C}_r \times p^r \times (1-p)^{n-r}$.
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c1}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c2}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c3}}\hspace{100px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c4R}}\hspace{100px}}~}$
$P(X = [2]) = [36] \times [0.8281] \times [0.00000005] = [0.00000143]$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c1}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c2}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c3}}\hspace{100px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c4R}}\hspace{100px}}~}$
$P(X = [3]) = [84] \times [0.753571] \times [0.00000053] = [0.00003364]$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c1}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c2}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c3}}\hspace{100px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c4R}}\hspace{100px}}~}$
$P(X = [4]) = [126] \times [0.68574961] \times [0.0000059] = [0.00051021]$
$P(X = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c0}}\hspace{35px}}~}) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c1}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c2}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c3}}\hspace{100px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c4R}}\hspace{100px}}~}$
$P(X = [5]) = [126] \times [0.62403215] \times [0.00006561] = [0.00515879]$
Therefore,
Hint: Add together the individual probabilities.
$P(1 < X \le 5) = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$
$P(1 < X \le 5) = [0.00570406]$