Question 402523: Two baseball players bat first and second in a lineup. The first batter has an on-base percentage of 0.34. The second batter has an on-base percentage of 0.46 if someone is on the base, but only 0.25 if the bases are empty. At the start of the game, what is the probability that neither player gets on base?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Two baseball players bat first and second in a lineup. The first batter has an on-base percentage of 0.34. The second batter has an on-base percentage of 0.46 if someone is on the base, but only 0.25 if the bases are empty. At the start of the game, what is the probability that neither player gets on base?
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P(not1st and not 2nd) = P(not 1st)*P(not 2nd | not 1st)
= 0.66*54
= 0.3564
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Cheers,
Stan H.
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