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?​

Answer 1

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.

Answer 2

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%