Question

A shipment of 20 similar laptop computers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives

Answer 1

Answer:

P(X=0) = 0.7158

P(X=1) = 0.2684

P(X=2) =0.0158

Step-by-step explanation:

Number of defective laptops = 3

Number of normal laptops = 17

If the school chooses 2 laptops from this lot, the number of defectives (X) could be 0, 1, or 2.

For zero defectives:

$P(X=0) = \frac{17}{20} *\frac{16}{19}\\P(X=0) = 0.7158$

For one defective:

$P(X=1) = \frac{17}{20} *\frac{3}{19}+\frac{3}{20} *\frac{17}{19}\\P(X=1) = 0.2684$

For two defectives:

$P(X=2) = \frac{3}{20} *\frac{2}{19}\\P(X=2) = 0.0158$