Two marksmen fire at a target simultaneously, the probability that Benita hits it but Jiri misses is 0.24 or 24%.
We have two marksmen, let consider two events,
A : Jiri hits the target
B : Benita hits the target
both the events are independent events.
Probability that benita hits the target, P(B) = 80% = 0.8
Probability ( Benita not hits the target) , P(Bᶜ) = 1 - P(B) = 0.2
Probability that Jiri hits the target, P(A)
= 70% =0.7
Probability ( Jiri not hits the target) , P(Aᶜ)
= 1 - P(A) = 0.3
We have to calculate the probability that Benita hits target but Jiri misses it, P( B and Aᶜ ). Now, the event Benita hits the target but jiri misses corresponds to ( B∩Aᶜ ) . Since, Benita and jiri hits the target independently, hence B and Aᶜ are also independent events.
Probability that Benita hits it but Jiri misses = P( B∩Aᶜ) = P(Aᶜ) P(B)
= 0.8 × 0.3 = 0.24 or 24%
Hence, required probability is 24%.
To learn more about independent events, refer :
https://brainly.com/question/14279727
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Complete question:
Two marksmen fire at a target simultaneously. Jiri hits the target 70% of the time and Benita hits the target 80% of the time. How do you determine the probability that Benita hits it but Jiri misses?