The probability that the team will win the second game given that they have already won the first game is 7/10
Event win in the 1st game is represented with A
Event win in the 2nd game is represented with B
In the question
i. The probability that they will win both games is P(AnB) = 42%
ii. The probability that they will win the first game is P(A) = 60%
The probability that the team will win the second game given that they have already won the first game is:
P(B/A) = P(BnA) / P(A)
P(B/A) = P(AnB) / P(A)
From the question
P(B/A) = 42% / 60%
P(B/A) = 0.42 / 0.60
P(B/A) = 7/10
In conclusion: The probability that the team will win the second game given that they have already won the first game is 7/10
See related question here https://brainly.com/question/23423241
