gemauert79

2022-09-17

A multiple choice test consists of 5 questions, each with 4 possible answers. An instructor will pass any student who can answer at least 1 question correctly. What is the probability that a student will pass using random guessing alone?
1. 0.7627
2. 0.2373
3. 0.8571
4. 0.25

hampiova76

Let x be number of questions correct. Here x follows binomial distribution with parameters $n=5$ and $p=\frac{1}{4}=0.25$
The pdf of binomial distribution is,
$P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}\left(1-p{\right)}^{n-x};x=0,1,2...$
The probability that a student will pass using random guessing alone is,
$P\left(X\ge 1\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)+P\left(X=5\right)\phantom{\rule{0ex}{0ex}}=\left(\begin{array}{c}5\\ 1\end{array}\right)\left(0.25{\right)}^{1}\left(1-0.25{\right)}^{5-1}+\left(\begin{array}{c}5\\ 2\end{array}\right)\left(0.25{\right)}^{2}\left(1-0.25{\right)}^{5-2}\phantom{\rule{0ex}{0ex}}+\left(\begin{array}{c}5\\ 3\end{array}\right)\left(0.25{\right)}^{3}\left(1-0.25{\right)}^{5-3}+\left(\begin{array}{c}5\\ 4\end{array}\right)\left(0.25{\right)}^{4}\left(1-0.25{\right)}^{5-4}\phantom{\rule{0ex}{0ex}}+\left(\begin{array}{c}5\\ 5\end{array}\right)\left(0.25{\right)}^{5}\left(1-0.25{\right)}^{5-5}\phantom{\rule{0ex}{0ex}}=0.3955+0.2637+0.0879+0.0147+0.0010\phantom{\rule{0ex}{0ex}}=0.7627$

Do you have a similar question?