Carcass wrote:
A shipment consists of 1,800 parts, some of which are defective. If a part is chosen from the shipment at random, the probability of it being defective is 1/9 the probability that it is not defective. How many of the parts in the shipment are defective?
(A) 90
(B) 120
(C) 180
(D) 200
(E) 900
The probability of it (the part) being defective is 1/9 the probability that it is not defectiveThis does
NOT mean P(getting a defective part) = 1/9
Let
x = the probability that the part is defective
So,
1 - x = the probability that the part is NOT defective
From the given information we can write: (probability of defective part) = (1/9)(probability of NON-defective part)
So we can write:
x = (1/9)(
1 - x)
To eliminate the fractions, multiply both sides by 9 to get: 9x = 1 - x
Add x to both sides: 10x = 1
Divide both sides by 10 to get:
x = 1/10So, P(electing a defective part) =
1/10In other words,
1/10 of all parts are defective.
If there are 1800 parts in total, then the number of defective parts =
1/10 of 1800 = 180
Answer: C
Cheers,
Brent