of the people who fished at clearwater park today, 49 had a fishing license, and 21 did not. of the people…

of the people who fished at clearwater park today, 49 had a fishing license, and 21 did not. of the people who fished at mountain view park today, 36 had a license, and 24 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 and the fisher chosen from mountain view had a license? do not round your answer.

of the people who fished at clearwater park today, 49 had a fishing license, and 21 did not. of the people who fished at mountain view park today, 36 had a license, and 24 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 and the fisher chosen from mountain view had a license? do not round your answer.

Answer

Explanation:

Step1: Calculate total fishers at Clearwater Park

Total fishers at Clearwater Park = 49 + 21 = 70.

Step2: Calculate probability of fisher from Clearwater without license

Probability of fisher from Clearwater without license = $\frac{21}{70}$.

Step3: Calculate total fishers at Mountain View Park

Total fishers at Mountain View Park = 36+24 = 60.

Step4: Calculate probability of fisher from Mountain View with license

Probability of fisher from Mountain View with license = $\frac{36}{60}$.

Step5: Calculate combined probability

Since the two events are independent, the combined probability is the product of the two probabilities. So the probability = $\frac{21}{70}\times\frac{36}{60}$. Simplify $\frac{21}{70}=\frac{3}{10}$ and $\frac{36}{60}=\frac{3}{5}$. Then $\frac{3}{10}\times\frac{3}{5}=\frac{9}{50}$.

Answer:

$\frac{9}{50}$