Question

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.

Answer 1

A= 0,2

B= 0,2

C= 0,4

D=0,2

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