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.

A = 0,2

B = 0,2

C = 0,4

D=0,2

Step-by-step explanation:

We know that only one team can win, so the sum of each probability of wining is one

P (A) + P (B) + P (C) + P (D) = 1

then we Know that the probability of Team A and B are the same, so

P (A) = P (B)

And that the the probability that either team A or team C wins the tournament is 0.6, so P (A) + Pc) = 0,6, then P (C) = 0.6-P (A)

Also, we know that team C is twice as likely to win the tournament as team D, so P (C) = 2 P (D) so P (D) = P (C) / 2 = (0.6-P (A)) / 2

Now if we use the first formula:

P (A) + P (B) + P (C) + P (D) = 1

P (A) + P (A) + 0.6-P (A) + (0.6-P (A)) / 2=1

0,5 P (A) + 0.9=1

0,5 P (A) = 0,1

P (A) = 0,2

P (B) = 0,2

P (C) = 0,4

P (D) = 0,2