Jaheim L.

asked • 12/31/19

What is the probability that the team will win the second game given that they have already won the first game?

The volleyball team has a double-header on Friday. The probability that they will win both games is 38%. The probability that they will win just the first game is 70%, What is the probability that the team will win the second game given that they have already won the first game?

John B.

tutor
There seems to be a problem with this question's wording. There are two possibilities for winning the first game: if you win it and the second or if you win it and lose the second. The sum of these two probabilities is the probability of winning the first game, regardless of the outcome of the second game (Law of Total Probability). That appears to be the information given in the question. There is a probability of 0.38 that they win both games, and there is a probability of 0.7 that they win the first and lose the second (that's how I interpret "win just the first game"). However, if we add these to find out the probability of winning the first game, regardless of the outcome of the second, then we get 0.7 + 0.38 = 1.08. But this is impossible because probabilities cannot be more than 1, or 100%. I'm thinking that whoever wrote the question didn't really intend this to be the meaning of the question. I'm guessing that they made a mistake in the wording and intended something much simpler: namely, that there is a 70% chance of winning the first game, period. If that is the intended meaning, then the answer is obtained by dividing 38 by 70, which is about 0.5428, or 54.28%.
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01/01/20

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Denise C. answered • 01/02/20

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It appears that either missing or misquoted wording in the question. Based on the information and how I interpret it, we know the following:


P(WW)=.38

P(1st win only, L 2nd game)=.7


Those are the only two outcomes when team has won the first game and those outcomes need to add up to 1, not 1.08.


Second interpretation: If you drop the word "only" So that ...


P(1st win) = .7,


then P(2nd win/given 1st win) = P(WW)/P(1st win)=.38/.7=.54


OR If they are not dependent events (As possibly interpreted in the problem wording), and P(any W) = .7, then P(2nd W) = .7


You may want to resubmit the problem with exact wording If there's something missing.

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