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

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step-by-step explanation:

step-by-step explanation:

sry but any answer choices