Thank you for visiting Bill and George go target shooting together Both shoot at a target at the same time Suppose Bill hits the target with probability 0 7. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Final answer:
You can find these probabilities using conditional probability. For part (a), the probability that George hit the target given that exactly one shot hit the target is approximately 0.222. For part (b), the probability that George hit the target given that the target is hit is approximately 0.348.
Explanation:
To solve these problems, we can use the rules of conditional probability. First, let's define our events: Let A be the event 'Bill hits the target' and B be the event 'George hits the target'.
(a) We want to find the probability that George hit the target given that exactly one shot hit the target. This means either Bill missed and George hit, or George missed and Bill hit. We calculate this as follows: Probability = P(George hit | One hit) = P(George hit AND Bill missed) / P(One hit). After substituting the given, we find the probability to be 0.4*0.3 / (0.4*0.3 + 0.7*0.6), which is approximately 0.222.
(b) We want to find the probability that George hit the target given that the target is hit. We calculate this as follows: Probability = P(George hit | Target is hit) = P(George hit AND Target is hit) / P(Target is hit). After substituting the given, we find the probability to be 0.4 / (0.7 + 0.4 - 0.7*0.4), which is approximately 0.348.
Learn more about Conditional Probability here:
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Rewritten by : Jeany
Answer: (a) [tex]\frac{2}{9}[/tex] (b) [tex]\frac{6}{41}[/tex]
Step-by-step explanation:
(a) P( Bill hitting the target) = 0.7 P( Bill not hitting the target) = 0.3
P( George hitting the target) = 0.4 P(George not hitting the target) = 0.6
Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3
= 0.54
Chances that George hit the target is = 0.4 x 0.3 = 0.12
So given that exactly one shot hit the target, probability that it was George's shot = [tex]\frac{0.12}{0.54}[/tex] = [tex]\frac{2}{9}[/tex] .
(b) The numerator in the second part would be the same as of (a) part which is 0.12.
The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.
Given that the target is hit,probability that George hit it = [tex]\frac{0.4\times 0.3}{0.7\times 0.6 +0.4\times 0.3+0.4\times 0.7}[/tex] = = [tex]\frac{6}{41}[/tex]
