Expert Answer to "At a local university, 54.3% of incoming first..."

Tucker Lewis

Answered question

2022-02-13

At a local university,

54.3 %

of incoming first year students have a PC. If 3 students are selected randomly, what are the probabilities that at least one has a computer?

vhudzeniy2n

Beginner2022-02-14Added 16 answers

Step 1
Let X denote number of students with computer in

n = 3

independent students, where each has a probability

p = 0.543

We denote this by:

X \sim Binomial (n = 3, p = 0.543)

Probability of x number of students with computer is:

P (X = x) = (\begin{matrix} n \\ x \end{matrix}) p^{x} {(1 - p)}^{n - x}

where

(\begin{matrix} n \\ x \end{matrix})

is combinatorial expression:

(\begin{matrix} n \\ x \end{matrix}) = \frac{n!}{x! (n - x)!}

P (X \geq 1), n = 3, p = 0.543

P (X \geq 1) = 1 - P (X < 1)

We first calculate:

P (X < 1) = P (0 \leq X < 1) = P (X = 0)

P (X = 0) = (\begin{matrix} 3 \\ 0 \end{matrix}) {(0.543)}^{0} {(1 - 0.543)}^{3} = 1 {(0.543)}^{0} {(0.457)}^{3} \approx 0.095444

P (X = 0) = 0.095444 \approx 0.095

P (X < 1) = 0.0954

P (X \geq 1) = 1 - P (X < 1) = 1 - 0.0954 = 0.9046

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