Question

Three friends (A, B, and C) will participate in a round-robin tournament in which each one plays both of the others. Suppose that P(A beats B) 0.4 P(A beats C) = 0.2 P(B beats C) = 0.8 and that the outcomes of the three matches are independent of one another. (a) What is the probability that A wins both her matches and that B beats C? (b) What is the probability that A wins both her matches? (c) What is the probability that A loses both her matches? (d) What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

Answer 1

We would be looking at the first set of questions - all 4 parts here:

Q1) a) We are given here that:
P(A beats B) = 0.4,
P(A beats C) = 0.2,
P(B beats C) = 0.8

a) Probability that A wins both the matches and that B beats C is computed using multiplication rule of probability here as: ( As the events are independent, we can use the multiplication rule here)

= P(A beats B)P(A beats C)P(B beats C) = 0.4*0.2*0.8 = 0.064

Therefore 0.064 is the required probability here.

b) Probability that A wins both the matches:

= P(A beats B)P(A beats C) = 0.4*0.2 = 0.08

Therefore 0.08 is the required probability here.

c) Probability that A loses both the matches is computed here as:

= (1 - P(A beats B))(1 - P(A beats C)) = 0.6*0.8 = 0.48

Therefore 0.48 is the required probability here.

d) Probability that each person wins one match is computed here as:

= P(A beats B, C beats A and B beats C) + P(A beats C, C beats B and B beats A)

= 0.4*(1 - 0.2)*0.8 + 0.2*(1 - 0.4)*(1 - 0.8)

= 0.4*0.8*0.8 + 0.2*0.6*0.2 = 0.28

Therefore 0.28 is the required probability here.