Question

use the frequency table to compute the following. probability of living in squirrel hill: ( p(a) = ) probability of living in mount oliver: ( p(b) = ) probability of living in squirrel hill and living in mount oliver: ( p(a \text{ and } b) = ) event a: living in squirrel hill

Explanation:

Step1: Calculate \(P(A)\)

Probability formula \(P = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\).
For \(P(A)\) (living in Squirrel Hill), number of favorable outcomes \(n(A)=60\), total number \(n = 182\).
\(P(A)=\frac{60}{182}=\frac{30}{91}\approx0.3297\)

Step2: Calculate \(P(B)\)

For \(P(B)\) (living in Mount Oliver), number of favorable outcomes \(n(B) = 56\), total number \(n=182\).
\(P(B)=\frac{56}{182}=\frac{28}{91}=\frac{4}{13}\approx0.3077\)

Step3: Calculate \(P(A\cap B)\)

Since a person can't live in both Squirrel Hill and Mount Oliver simultaneously (mutually - exclusive events), \(n(A\cap B) = 0\).
Using the formula \(P(A\cap B)=\frac{n(A\cap B)}{n}\), we get \(P(A\cap B)=\frac{0}{182}=0\)

Answer:

\(P(A)=\frac{30}{91}\), \(P(B)=\frac{4}{13}\), \(P(A\cap B) = 0\)