Question 363026: * The probability that a projector will have a defective lens is 0.12, the probability that it will have a crack defect is 0.29, and the probability that it has both defects is 0.07.
iii) What is the probability that a newly manufactured projector has exactly one of the two defects?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let x = lens defect
let y = crack defect
p(x) = .12
p(y) = .29
p(x+y) = .07
the probability of p(x) without p(x+y) should be equal to p(x) - p(x + y) = .12 - .07 = .05
the probability of p(x) without p(x+y) should be equal to p(y) - p(x + y) = .29
we'll examine these with a hypothetical situation to see what's happening.
the probability of x occurring is .12.
this means that, if the probabilities hold, and you sell 1000 products, then:
12% of them will have a lens defect.
29% of them will have a crack defect.
12% of 1000 = 120
29% of 1000 = 290
total defective products are therefore 390.
of these .07 * 1000 = 70 have both a crack defect and a lens defect.
to find how many products have only a lens defect or only a crack defect, you have to subtract 70 from both to get:
50 have only a lens defect.
220 have only a crack defect.
50 / 1000 = .05
220 / 1000 = .22
.05 = .12 - .05 which is the same as p(x) - p(x + y)
.22 = .29 - .05 which is the same as p(y) - p(x + y)
the formulas look good.
the question is, of the total products sold, how many have only a lens defect or only a crack defect.
50 have only a lens defect and 220 have only a crack defect which makes a total of 270 with only a lens defect or only a crack defect.
270 / 1000 = .27 which means that 27% of the time this occurs which means that the probability of only a lens defect or only a crack defect = .27.
that should be your answer.
to go one step further, let's take the probability of a product having a defect.
this could be either lens only or crack only or both.
the equation for that is:
p(x or y) = p(x) + p(y) - p(x + y)
that comes out to be .12 + .29 - .07 = .34
34% of the time a product will have a defect.
it will be either a lens only or a crack only or a lens and a crack.
you have to subtract p(x + y) because (x + y) is both a memmber of x and a member of y.
if you want to know the probability that it will have a lens only or a crack only, then you have to subtract products that have a lens defect and a crack defect.
the formula becomes p(x or y) - p(x + y) = p(x) + p(y) - p(x + y) - p(x + y) which becomes:
p(x or y) - p(x + y) = p(x) + p(y) - 2*p(x + y)
so we have:
p(x or y) - p(x + y) = .12 + .29 - 2*.07 = .41 - .14 = .27
the hypothetical numbers confirm the probabilities as being correct.
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