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
Algebra ->
Probability-and-statistics
-> 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
Log On
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) (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) =
P(at least one occurs) = (given)
So
P(Q occurs but S does not) = y(1-x)
P(Q occurs but S does not) = (given)
So
Therefore we have this system:
Simplifying the first:
Simplifying the second:
So the simplified system is:
Multiply the first equation through by -3:
Adding the equations term by term:
That's all that was asked for. However
it you had been asked for the probability
of Q you would substitute into
So and
Edwin