In a certain city, 63.9% of the voters are Democrats and 36.1% of the voters are
ID: 3010779 • Letter: I
Question
In a certain city, 63.9% of the voters are Democrats and 36.1% of the voters are Republicans. In the last election 40.1% of the Democrats, and 43.8% of the Republican voted. a) If the citizen who voted in the last election is randomly selected, what is the probability that they are a Democrat? b) If a citizen who is a Republican is randomly selected, what is the probability that they did not vote in the last election? c) if the citizen is selected at random, what is the probability that they are a republican and they voted?
Explanation / Answer
Probability of being a Democrat, P(D) = 0.639
Probability of being a Republican, P(R) = 0.361
% of Democrat who votes last time, d = 40.1 %
% of republican who voted last time, r = 43.8%
a) If the citizen who voted in the last election is randomly selected, what is the probability that they are a Democrat?
Probability that the citizen voted last time= dP(D)+rP(R) = 0.401*0.639+0.438*361 = 0.414
Probability that a Democrat voted last time = dP(D) / {dP(D)+rP(R)} = 0.401*0.639 / {0.401*0.639+0.438*361} = 0.618
b) Probability that a Republican did not vote last time = (1-r) = 1-0.438 = 0.562
c) Probabilty that a citizen is Republican and he voted last time = rP(r) = 0.438*0.361 = 0.158