Question

According to exit polls from the 2000 presidential election, the probability of a voter identifying as gay (including lesbians) was P(G) = 0.04. The probability of voting for Bush, given that a voter was gay, was P(B given G) = 0.25, and the probability of voting for Bush, given that a voter was not gay, was P(B given not G) = 0.50.
Find P(G intersection B)
Find P(G' intersection B) Find P(B), the overall probability of voting for Bush, keeping in mind that a voter was either gay and voted for Bush or not gay and voted for Bush.

Answer 1

Answer:

P(G∩B)=0.01

P(G'∩B)=0.48

P(B)=0.49

Step-by-step explanation:

The given probabilities are

P(G)=0.04

P(B/G)=0.25

P(B/G')=0.5

P(G intersection B)=P(G∩B)=?

According to definition of conditional probability

P(B/G)=P(G∩B)/P(G)

So,

P(G∩B)=P(G)*P(B/G)

P(G∩B)=0.04*0.25

P(G∩B)=0.01

Thus, P(G intersection B)=P(G∩B)=0.01.

Now, P(G' intersection B)=P(G'∩B)=?

According to definition of conditional probability

P(B/G')=P(G'∩B)/P(G')

So,

P(G'∩B)=P(G')*P(B/G')

P(G')=1-P(G)=1-0.04=0.96

P(G'∩B)=0.96*0.5

P(G'∩B)=0.48

Thus, P(G' intersection B)=P(G'∩B)=0.48.

Now, P(B)=?

We know a voter is either gay and voted for Bush or not gay and voted for Bush so,

P(B)=P(G∩B)+P(G'∩B)

P(B)=0.01+0.48

P(B)=0.49

Thus, the overall probability of voting for Bush P(B)=0.49.