SOLUTION: Suppose Q and S are independent events such that the probability that at least one of them occurs is 1/3 and the probability that Q occurs but S does not occur is 1/9. What is the

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Question 252131: Suppose Q and S are independent events such that the probability that at least one of them occurs is 1/3 and the probability that Q occurs but S does not occur is 1/9. What is the probability of S?
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose Q and S are independent events such that the probability that at least one of them occurs is 1/3 and the probability that Q occurs but S does not occur is 1/9. What is the probability of S?

Let P(S occurs) = x 
Let P(Q occurs) = y

Then 

P(S does not occur) = 1 - x
P(Q does not occur) = 1 - y
P(neither occurs) = (1 - x)(1 - y)


P(at least one occurs) = 1 - P(neither occurs) = 1+-+%281+-+x%29%281+-+y%29

P(at least one occurs) = 1%2F3  (given)

So  1+-+%281+-+x%29%281+-+y%29+=+1%2F3

P(Q occurs but S does not) = y(1-x)

P(Q occurs but S does not) = 1%2F9  (given)

So y%281-x%29=1%2F9

Therefore we have this system:

system%281+-+%281+-+x%29%281+-+y%29+=+1%2F3%2C+y%281-x%29=1%2F9%29

Simplifying the first:

1+-+%281+-+x%29%281+-+y%29+=+1%2F3

1+-+%281+-+y+-x+%2B+xy%29+=+1%2F3

1+-+1+%2B+y%2B+x+-+xy+=+1%2F3

y%2B+x+-+xy+=+1%2F3
3y+%2B+3x+-+3xy+=+1

Simplifying the second:

y%281-x%29=1%2F9
y-xy=1%2F9
9y-9xy=1

So the simplified system is:

system%283y%2B3x-3xy=1%2C+9y-9xy=1%29

Multiply the first equation through by -3:

system%28-9y-9x%2B9xy=-3%2C+9y-9xy=1%29

Adding the equations term by term:

-9x=-2

x=2%2F9

That's all that was asked for.  However
it you had been asked for the probability
of Q you would substitute x=2%2F9 into

9y-9xy=1

9y-9%282%2F9%29y=1

9y-cross%289%29%282%2Fcross%289%29%29y=1


9y-2y=1

7y=1

y=1%2F7

So %22P%28S%29%22+=2%2F9 and %22P%28Q%29%22=1%2F7

Edwin