A probability question based on batting averages,

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SUMMARY

The discussion centers on calculating the probability that neither Batter A, with a batting average of .320, nor Batter B, with a batting average of .280, gets a hit during an inning. The correct approach involves determining the probabilities of each batter not getting a hit, which are 0.680 for Batter A and 0.720 for Batter B. Multiplying these probabilities yields a combined probability of approximately 0.4896, or 48.96%, that both batters will miss. The calculations were confirmed as accurate by other participants in the discussion.

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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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