Question

In a certain town in the united states, 40% of the population are democrats and 60% are republicans. The municipal government has proposed making gun ownership illegal in the town. It is known that 75% of democrats and 30% of republicans support this measure. If a resident of the town is selected at random.

(a) what is the probability that they support the measure?
(b) if the selected person does support the measure what is the probability the person is a democrat?
(c) if the person selected does not support the measure, what is the probability that he or she is a democrat?

Answer 1

(a) 0.48, (b) 0.625, (c) 0.1923

Explanation:

First define

Probability a person is a democrat: P(D) = 0.4

Probability a person is a republican: P(R) = 0.6

Probability a person support the measure given that the person is a democrat: P(SM | D) = 0.75

Probability a person support the measure given that the person is a republican: P(SM | R) = 0.3

Now for the Theorem of total probabilities we have

(a) P(SM) = P(SM | D)P(D)+P(SM | R)P(R) = (0.75)(0.4)+(0.3)(0.6) = 0.48

and for the Bayes' Formula we have

(b) P(D | SM) = P(SM | D)P(D)/[P(SM | D)P(D)+P(SM | R)P(R)] = (0.75)(0.4)/0.48 = 0.625

Now let SMc be the complement of support the measure, i.e.,

P(SMc | D) = 0.25 : Probability a person does not support the measure given that the person is a democrat

P(SMc | R) = 0.7: Probability a person does not support the measure given that the person is a republican,

and also for the Bayes' Formula we have

(c) P(D | SMc) = P(SMc | D)P(D)/[P(SMc | D)P(D)+P(SMc | R)P(R)] = (0.25)(0.4)/[(0.25)(0.4)+(0.7)(0.6)] = 0.1/(0.52)=0.1923