Question 884662: A batch of 50 parts contains six defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective?
If the experiment is repeated, with replacement, what is the probability that both parts are defective?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! batch of 50 contains 6 defective.
p(defective) = 6/50
draw 2 with replacement.
p(dd) = 6/50 * 6/50
draw 2 without replacement.
p(dd) = 6/50 * 5/49
in the first case, the probability of drawing the second defective is independent on whether you draw a defective on the first draw or not.
in the second case, the probability of drawing the second defective is dependent on whether you draw a defective on the first draw or not.
the formulas for both these considerations can be shown as follows:
let a be the event of drawing a defective on the first draw.
p(a) = 6/50
let b be the event of drawing a defective on the first draw as well.
p(b) = 6/50
if they are independent:
p(a and b) = p(a) * p(b) = 6/50 * 6/50
if they are dependent:
p(b given a) = p(a and b) / p(a).
solve for p(a and b) and this formula becomes:
p(a and b) = p(a) * p(b given a)
we need to determine what p(b given a) is.
p(b given a) means the probability of getting b given that you already got a.
you start with 6 defectives out of 50
you draw a part and it is defective.
you are left with 5 defectives out of 49.
p(b given a) is therefore equal to 5/49.
the formula of p(a and b) = p(a) * p(b given a) becomes:
p(a and b) = p(a) * p(b given a) = 6/50 * 5/49
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