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?

Answers

Answer 1

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


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Step-by-step explanation:

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Answers

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Answers

9514 1404 393

Answer:

  see attached

Step-by-step explanation:

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__

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