SOLUTION: 3. The first AUM baseball player coming to bat in an inning gets a hit 25% of the time. The second player coming to bat in an inning gets a hit 32% of the time. Consider their at-b
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-> SOLUTION: 3. The first AUM baseball player coming to bat in an inning gets a hit 25% of the time. The second player coming to bat in an inning gets a hit 32% of the time. Consider their at-b
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Question 1012225: 3. The first AUM baseball player coming to bat in an inning gets a hit 25% of the time. The second player coming to bat in an inning gets a hit 32% of the time. Consider their at-bats together, assuming that the outcomes of these two at-bats are independent of each other. What is the probability that neither of them gets a hit? Answer by Theo(13342) (Show Source):
second player gets a hit 32% of the time.
this means he doesn't get a hit 68% of the time.
.75 * .68 = .51.
neither will get a hit 51% of the time.
the total probabilities are:
both get a hit = .25 * .32 = .08
first gets a hit and second doesn't = .25 * .68 = .17
first doesn't get a hit and second does = .75 * .32 = .24
both gon't get a hit = .75 * .68 = .51
add up all the probabilities and you should get 1.
.08 + .17 + .24 + .51 = 1
