Question 1195929
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There are of course many ways to set up and solve this problem; and the standard methods taught in typical textbooks are not always the easiest way.<br>
The numbers in this problem lend themselves to a quick and easy solution using a non-standard method.<br>
We are given...
(1) P(A) = P(B) = x  (Teams A and B have equal probability of winning)
(2) P(D) = y; P(C) = 2y  (Team C is twice as likely to win as team D)
(3) P(A)+P(C) = x+2y = 0.6  (The probability that either A or C wins is 0.6)<br>
Since the sum of all the probabilities is 1, we also have...
(4) x+x+y+2y = 2x+3y = 1<br>
So we have two equations that we can solve to find the solution to the problem.  We could of course use the standard methods of substitution or elimination; but there is what I think is a much easier path to the answer.<br>
(4) tells us 2x+3y=1; (3) tells us x+2y=0.6.  Comparing those two gives us<br>
(5) x+y=0.4<br>
Then comparing (5) and (3) gives us y=0.2.<br>
And after that everything is easy:
P(D) = y = 0.2
P(C) = 2y = 0.4
That leaves P(A)+P(B) = 0.4; and since P(A)=P(B), P(A)=P(B)=0.2.<br>
ANSWERS:
P(A) = 0.2
P(B) = 0.2
P(C) = 0.4
P(D) = 0.2<br>