In a certain town 391 of the voters are Democrats and 609 of

In a certain town, 39.1% of the voters are Democrats and 60.9% of the voters are Republicans. In the last election 30.8% of the Democrats, and 39.6% of the Republicans voted.

a.) If a citizen who is a Republican is randomly selected, what is the probability that they did not vote in the last election?

Solution

Solution: Let D₁, D₂, E₁, E₂,R, are defined as:

D₁ = selected voters are Democrats in first eliction = 39.1%.

D₂ = selected voters are Democrats in last eliction = 30.8%.

E₁= selected voters are Republicans in first eliction = 60.9%.

E₂ = selected voters are Republicans in the last eliction = 39.6%.

R = selected voters is Republicans

If a citizen who is a Republican is selected randomly and they did not vote in the last election means that he is selected from first ellection.

Thus P(R₁) =60.9/100 = 0.609.

Thus P(R₂) =39.6/100 = 0.396.

P(R/E₁) =

Use Baye\'s Theorem

Probability of a citizen being selected is a Republican in first election = 60.9/100.= 0.609.