Question 1195929: Four teams A, B, C, and D compete in a tournament, and exactly one of them will
win the tournament. Teams A and B have the same chance of winning the tournament. Team C is
twice as likely to win the tournament as team D. The probability that either team A or team C wins
the tournament is 0.6. Find the probabilities of each team winning the tournament.
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
P(A) = x
P(B) = x
P(C) = 2y
P(D) = y
These are the probabilities of each team winning the tournament
The values are such that  and 
Those probabilities add to 1 since they represent 100% of all possible cases.
P(A)+P(B)+P(C)+P(D) = 1
x+x+2y+y = 1
2x+3y = 1
The probability of team A or C winning is 0.6, so,
P(A or C) = P(A) + P(C) ... mutually exclusive
P(A or C) = 0.6
P(A) + P(C) = 0.6
x+2y = 0.6
x = -2y+0.6
Apply substitution
2x+3y = 1
2( x )+3y = 1
2( -2y+0.6 )+3y = 1
-4y+1.2+3y = 1
-y+1.2 = 1
-y = 1-1.2
-y = -0.2
y = 0.2
Therefore, P(C) = 2*y = 2*0.2 = 0.4 and P(D) = y = 0.2
And,
x = -2y+0.6
x = -2(0.2)+0.6
x = -0.4+0.6
x = 0.2
which means P(A) = P(B) = x = 0.2
Summary:
P(A) = 0.2
P(B) = 0.2
P(C) = 0.4
P(D) = 0.2
Check:
P(A)+P(B)+P(C)+P(D) = 1
0.2+0.2+0.4+0.2 = 1
1 = 1
and,
P(A or C) = P(A) + P(C) = 0.2 + 0.4 = 0.6
The answers are confirmed.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
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.
The numbers in this problem lend themselves to a quick and easy solution using a non-standard method.
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)
Since the sum of all the probabilities is 1, we also have...
(4) x+x+y+2y = 2x+3y = 1
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.
(4) tells us 2x+3y=1; (3) tells us x+2y=0.6. Comparing those two gives us
(5) x+y=0.4
Then comparing (5) and (3) gives us y=0.2.
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.
ANSWERS:
P(A) = 0.2
P(B) = 0.2
P(C) = 0.4
P(D) = 0.2
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