Formulaic Maths Problems
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Find $P(3 \le X \le 8)$, where $X$ is a Binomial variable with parameters $n = 9$, $p = 0.84$.
Find $1 - p$.
$1 - p = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~}$
$1 - p = [0.16]$
Hint: For a Binomial distribution, $P(X = r) = {}^n \mathrm{C}_r \times p^r \times (1-p)^{n-r}$.
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c1}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c2}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c3}}\hspace{100px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c4R}}\hspace{100px}}~}$
$P(X = [3]) = [84] \times [0.592704] \times [0.00001678] = [0.00083529]$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c1}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c2}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c3}}\hspace{100px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c4R}}\hspace{100px}}~}$
$P(X = [4]) = [126] \times [0.49787136] \times [0.00010486] = [0.00657791]$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c1}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c2}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c3}}\hspace{100px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c4R}}\hspace{100px}}~}$
$P(X = [5]) = [126] \times [0.41821194] \times [0.00065536] = [0.034534]$
$P(X = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c0}}\hspace{35px}}~}) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c1}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c2}}\hspace{100px}}~} \times \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c3}}\hspace{100px}}~} = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br3c4R}}\hspace{100px}}~}$
$P(X = [6]) = [84] \times [0.35129803] \times [0.004096] = [0.12086901]$
$P(X = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c0}}\hspace{35px}}~}) = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c1}}\hspace{100px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c2}}\hspace{100px}}~} \times \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c3}}\hspace{100px}}~} = \class{inputBox step6}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br4c4R}}\hspace{100px}}~}$
$P(X = [7]) = [36] \times [0.29509035] \times [0.0256] = [0.27195526]$
$P(X = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br5c0}}\hspace{35px}}~}) = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br5c1}}\hspace{100px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br5c2}}\hspace{100px}}~} \times \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br5c3}}\hspace{100px}}~} = \class{inputBox step7}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br5c4R}}\hspace{100px}}~}$
$P(X = [8]) = [9] \times [0.24787589] \times [0.16] = [0.35694128]$
Therefore,
Hint: Add together the individual probabilities.
$P(3 \le X \le 8) = \class{inputBox step8}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$
$P(3 \le X \le 8) = [0.79171275]$