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

Question

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 just the first game is 60%, What is the probability that the team will win the 2nd game given that they have already won the first game?​

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Verity 6 months 2021-07-17T07:25:52+00:00 2 Answers 16 views 0

Answers ( )

    0
    2021-07-17T07:27:17+00:00

    Answer:

    24%

    Step-by-step explanation:

    assign a variable x:

    Find the average and solve for x by isolating the variable:

    (x+60%)÷2=42%

    multiply on both sides, left side cancels, multiply right side

    (x+60%)÷2 *2= 42%*2

    x+60% = 84%

    subtract on both sides, left side cancels, subtract right side

    x+60% – 60%=84% – 60%

    x=24%

    0
    2021-07-17T07:27:25+00:00

    This problem fits the conditional probability formula very well. The formula is P(B|A) = P(B ∩ A)/P(A). If event A is winning the first game, and event B is winning the second, then P(B ∩ A) = 0.44, and P(A) = 0.6. So P(B|A) is obtained by dividing 0.44 by 0.6, which is about 0.733.

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