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of the people who fished at Clearwater park today 36 had a fishing license and four did not if the people who fished at Mountain View Park today 35 had a license in 15 did not no one fished at both parks suppose that one Fisher from each Park is chosen at random what is the probability that the Fisher chosen from Clearwater did not have a license in the Fisher chosen from Mountain View had a license​

User Ghasem Deh
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Answer:

There are a couple of ways to approach this problem, but one possible method is to use conditional probability. To find the probability that the Fisher chosen from Clearwater did not have a license and the Fisher chosen from Mountain View had a license, we can use the formula:

P(A and B) = P(A) * P(B|A)

Where A is the event that the Fisher chosen from Clearwater did not have a license and B is the event that the Fisher chosen from Mountain View had a license.

To find P(A), we can use the total number of Fishers at Clearwater who did not have a license divided by the total number of Fishers at Clearwater. So, P(A) = 4 / 40 = 1/10

To find P(B|A), we can use the total number of Fishers at Mountain View who had a license divided by the total number of Fishers at Mountain View. So, P(B|A) = 35 / 50 = 7/10

Then we can substitute these values into the formula:

P(A and B) = P(A) * P(B|A) = (1/10) * (7/10) = 7/100

So the final probability that the Fisher chosen from Clearwater did not have a license in the Fisher chosen from Mountain View had a license is 7/100.

Explanation:

User Gzbwb
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