A probability question based on batting averages,

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Batter A has a batting average of .320, while Batter B has a batting average of .280. To find the probability that neither gets a hit, the probabilities of each not getting a hit are calculated as 0.680 for A and 0.720 for B. Multiplying these probabilities gives a combined probability of approximately 0.4896, or 48.96%. The calculations were confirmed as correct, providing reassurance for the original poster's understanding. This illustrates the application of independent probabilities in batting averages.
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Batter A has a batting average (probability of getting a hit) of .320. Batter B has a
batting average of .280. If both of them come to bat during an inning, calculate the
probability that neither one gets a hit. Assume that A getting a hit is independent of B
getting a hit.

My attempt:

I said that .320 is basically 320/1000 and that's 8/25 and .280 is 7/25. I said that the probability of them not getting a hit is 17/25 for Batter A and 18/25 for batter B. And I multiplied both of em to get 48.96% chance of both of em missing. Is this right? I'm not really a math type person but this question is for one of my science classes. Thanks for any help :)
 
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shadedude123 said:
Batter A has a batting average (probability of getting a hit) of .320. Batter B has a
batting average of .280. If both of them come to bat during an inning, calculate the
probability that neither one gets a hit. Assume that A getting a hit is independent of B
getting a hit.

My attempt:

I said that .320 is basically 320/1000 and that's 8/25 and .280 is 7/25. I said that the probability of them not getting a hit is 17/25 for Batter A and 18/25 for batter B. And I multiplied both of em to get 48.96% chance of both of em missing. Is this right? I'm not really a math type person but this question is for one of my science classes. Thanks for any help :)

100% correct.
 
Thanks a lot :)
 

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