The baseball team has a double-header on Saturday. The probability that they will win both games is 42%. The probability that they will win the first game is 60%, What is the probability that the team will win the second game given that they have already won the first game?

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 link

answer: idk

step-by-step explanation:

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11