Question

Of the people who fished at Clearwater Park today, 48 had a fishing license, 32and did not. Of the people who fished at Mountain View Park today, 72 had a license, and 18 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

Answer:

The probability that the fisher chosen from Clearwater did not have a license and the fisher chosen from Mountain View had a license is 0.32.

Step-by-step explanation:

Denote the events as follows:

X = a fisher at Clearwater Park had a fishing license

Y = a fisher at Mountain View Park had a fishing license

The two events are independent.

The information provided is:

n (X) = 48

n (X') = 32

n (Y) = 72

n (Y') = 18

Then,

N (X) = n (X) + n (X')

        = 48 + 32

        = 80

N (Y) = n (Y) + n (Y')

        = 72 + 18

        = 90

Compute the probability that the fisher chosen from Clearwater did not have a license and the fisher chosen from Mountain View had a license as follows:

$P(X'\cap Y)=P(X')\times P(Y)$

                 $=\frac{n(X')}{N(X)}\times \frac{n(Y)}{N(Y)} \\\\=\frac{32}{80}\times\frac{72}{90}\\\\=0.32$

Thus, the probability that the fisher chosen from Clearwater did not have a license and the fisher chosen from Mountain View had a license is 0.32.