Questions: Of the people who fished at Clearwater Park today, 30 had a fishing license, and 20 did not. Of the people who fished at Mountain View Park today, 32 had a license, and 8 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 had a license and the fisher chosen from Mountain View did not have a license?

Solution Steps

First, we need to calculate the total number of fishers at each park.

• Clearwater Park:

  • Fishers with a license: 30
  • Fishers without a license: 20
  • Total fishers = 30 + 20 = 50

• Mountain View Park:

  • Fishers with a license: 32
  • Fishers without a license: 8
  • Total fishers = 32 + 8 = 40

Next, we calculate the probability of each event occurring:

• Probability that the fisher chosen from Clearwater had a license: \[ P(\text{License at Clearwater}) = \frac{30}{50} = 0.6 \]

• Probability that the fisher chosen from Mountain View did not have a license: \[ P(\text{No License at Mountain View}) = \frac{8}{40} = 0.2 \]

Since the events are independent (choosing one fisher from each park), we multiply the probabilities to find the combined probability:

\[ P(\text{License at Clearwater and No License at Mountain View}) = P(\text{License at Clearwater}) \times P(\text{No License at Mountain View}) = 0.6 \times 0.2 = 0.12 \]

Final Answer