[Math] Probability Modem is Defective

combinatoricsprobability

A store has 80 modems in its inventory, 30 coming from Source A and
the remainder from Source B. Of the modems from Source A, 20% are
defective. Of the modems from Source B, 8% are defective. Calculate
the probability that exactly two out of a random sample of five modems
from the store’s inventory are defective.

(A) 0.010 (B) 0.078 (C) 0.102 (D) 0.105 (E) 0.125

A: Modem is from source A, P(A) = 30/80

B: Modem is from source B, P(B) = 50/80

To have 2 defective modems we could have AA (1st and 2nd defective modem from A), AB (1st defective modem from A, 2nd from B), BA, BB.

The probability of choosing each modem and each being defective is thus:

P(exactly 2 defective) = $\frac{30}{80}\frac{29}{79}(0.2^2)+\frac{30}{80}\frac{50}{79}(0.2*.08)*2+\frac{50}{80}\frac{49}{79}(0.08^2) = 0.016$

My answer didn't match any of the choices, what flaw am I making in my thinking?

Edit: I know the correct answer, and saw a solution using combinations, but I want to know if there is another way to solve the problem taking an approach similar to mine.

The binomial approach the other posters used led to the wrong result as well. What mistake is being made?

Best Answer

Good grief! Sorry, misread the problem. With your prompting (and some coffee with lunch), I think I have it right now.

Altogether, there are 10 defectives among 80. You are selecting five out of those 80, and you want the probability of getting exactly two defectives. This is hypergeometric.

$$P(X = 2) = \frac{{10 \choose 2}{70 \choose 3}}{{80 \choose 5}} = 0.1024667.$$

In R statistical software:

 dhyper(2, 10, 70, 5)
 ## 0.1024667
 choose(10,2)*choose(70,3)/choose(80,5)
 ## 0.1024667
Related Question