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Find $P(X \le 2)$, where $X$ is a Binomial variable with parameters $n = 3$, $p = 0.1$.
Find $1 - p$.
$1 - p = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{A1}}\hspace{54px}}~}$ $1 - p = 0.9$
Hint: For a Binomial distribution, $P(X = r) = {}^n \mathrm{C}_r \times p^r \times (1-p)^{n-r}$.
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c1}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c2}}\hspace{100px}}~} \times \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c3}}\hspace{100px}}~} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br0c4R}}\hspace{100px}}~}$ $P(X = 0) = 1 \times 1 \times 0.729 = 0.729$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c1}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c2}}\hspace{100px}}~} \times \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c3}}\hspace{100px}}~} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br1c4R}}\hspace{100px}}~}$ $P(X = 1) = 3 \times 0.1 \times 0.81 = 0.243$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c1}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c2}}\hspace{100px}}~} \times \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c3}}\hspace{100px}}~} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Br2c4R}}\hspace{100px}}~}$ $P(X = 2) = 3 \times 0.01 \times 0.9 = 0.027$
Therefore, Hint: Add together the individual probabilities.
$P(X \le 2) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{CR}}\hspace{100px}}~}$ $P(X \le 2) = 0.999$