Question

Of the people who fished at Clearwater Park today,

56 had a fishing license, and

14 did not. Of the people who fished at Mountain View Park today,

54 had a license, and 6 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?

Answer 1

For the people who fished at Clearwater Park today:

The total number of outcomes = 56+14 = 70

For the people who fished at Mountain View Park today:

Total number of outcomes = 54+6 = 60

Probability = Number of favourable outcome/Total number of outcomes.

P(Fisher chosen from Clearwater did not have lisense) = 14/(56 + 14) = 1/5

P(Fisher chosen from Mountain view had a lisense) = 54/(54 + 6) = 9/10

So the probability that the fisher chosen from Clearwater did not have a license and the fisher chosen from Mountain View had a license is

= P(fisher chosen from Clearwater did not have a license) × P(fisher chosen from Mountain View had a license)

= (1/5) × (9/10)

= 9/50

= 0.18

Therefore the required probability is 0.18