Question

Gender and political party In January$2017$, $52\%$of U.S. senators were Republicans and

the rest were Democrats or Independents. Twenty-one percent of the senators were

females, and $47\%$of the senators were male Republicans. Suppose we select one of these

senators at random. Define events R: is a Republican and M: is male.

a. Find P(R ∪ M). Interpret this value in context.

b. Consider the event that the randomly selected senator is a female Democrat or

Independent. Write this event in symbolic form and find its probability.

Short answer

(a) The probability of those who are either republicans or male is$0.84$.

(b) The symbolic form is$P\left(F\cup I\right)$and the probability$0.16$.

Step-by-step solution

Part (a) Step 1: Given Information

We are given the values of those who are republicans and those who are female democrats and those who are male and republicans both, and we have to find the probability of those who are either republicans or male.

Part (a) Step 2: Explanation

Applying the union rule of probability

which is$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)+P(A\cap B)$,

Here, P (R)$=0.52$, P (M)$=1-P\left(F\right)=1-0.21=0.79$and P$\left(R\cap M\right)=0.47$

Put all the values in the rule.

We get $P\left(R\cup M\right)=0.52+0.79-0.47=0.84$

Hence, the probability of those who are republican or male is$0.84$.

Part (b) Step 1: Given Information

We are given the values of those who are republicans and those who are female democrats and those who are male and republican both, and we have to find out the probability of those who are female or independent.

Part (b) Step 2: Explanation

Applying the union rule of probability,

which is$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)+P(A\cap B)$

Here, the probability of students who are female or independent is denoted by$P\left(F\cup I\right)$. Now, to find its value,

P(F)$=0.21$, P(I)$=1-P\left(R\right)=1-0.52=0.48$

and P$(F\cap I)=0.53$

Put all the values in the union rules.

We get$P(F\cup I)=0.21+0.48-0.53=0.16$.

Hence, the probability is$0.16$.