Question: "A collection of 22 laptops includes 6 defective..."

Sabrina Waller

Answered question

2022-02-09

A collection of 22 laptops includes 6 defective laptops. If a sample of 3 laptops is randomly chosen from the collection, what is the probability that at least one laptop in the sample will be defective?

Answer & Explanation

Keely House

Beginner2022-02-10Added 13 answers

Step 1
The likelihood that a laptop is broken is $(\frac{6}{22})$
The likelihood that a laptop won't have a problem is $(\frac{16}{22})$
The probability that at least one laptop is defective is given by:
$P (1 defective) + P (2 defective) + P (3 defective)$
as this probability is cumulative. Let X be the number of laptops found to be defective.
$P (X = 1) = (3 choose 1) {(\frac{6}{22})}^{1} \times {(\frac{16}{22})}^{2} = 0.43275$
$P (X = 2) = (3 choose 2) {(\frac{6}{22})}^{2} \times {(\frac{16}{22})}^{1} = 0.16228$
$P (X = 3) = (3 choose 3) {(\frac{6}{22})}^{3} = 0.02028$
(Sum up all of the probabilities)
$= 0.61531 \approx 0.615$

hotlegsaprile2u

Beginner2022-02-11Added 21 answers

Step 1

P (X \geq 1), N = 22, K = 6, n = 3

P (X \geq 1) = P (1 \leq X \leq 3) = 1 - P (X \leq 0) = 1 - (P (X = 0))

P (X = 0) = \frac{(\begin{matrix} K \\ 0 \end{matrix}) (\begin{matrix} N - K \\ n - 0 \end{matrix})}{(\begin{matrix} N \\ n \end{matrix})}

= \frac{(\begin{matrix} 6 \\ 0 \end{matrix}) (\begin{matrix} 22 - 6 \\ 3 - 0 \end{matrix})}{(\begin{matrix}  \end{matrix})} = \frac{(\begin{matrix} 6 \\ 0 \end{matrix}) (\begin{matrix} 16 \\ 3 \end{matrix})}{(\begin{matrix} 22 \\ 3 \end{matrix})}

= \frac{1 \times 560}{1540} = \frac{560}{1540} = \frac{4}{11}

= 0.363636

Answer:

= 1 - (0.363636) = 0.636364

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