Question

identifying two independent events
brown law firm collected data on the transportation choices of its employees for their morning commute. the table shows the percentages of the type of transportation of the male and female employees.
consider the following events:
a: the employee is male.
b: the employee is female.
c: the employee takes public transportation.
d: the employee takes his/her own transportation.
e: the employee takes some other method of transportation.
which two events are independent?
a and c
a and d
b and d
b and e

Explanation:

Step1: Recall the definition of independent events

Two events \(A\) and \(B\) are independent if \(P(A\cap B)=P(A)\times P(B)\).

Step2: Calculate probabilities for each option

Option A and C

\(P(A)=\frac{36}{60} = 0.6\), \(P(C)=\frac{20}{60}=\frac{1}{3}\approx0.33\), \(P(A\cap C)=\frac{12}{60} = 0.2\), and \(P(A)\times P(C)=0.6\times\frac{1}{3}= 0.2\). Since \(P(A\cap C)=P(A)\times P(C)\), \(A\) and \(C\) are independent.

Option A and D

\(P(A)=\frac{36}{60}=0.6\), \(P(D)=\frac{30}{60}=0.5\), \(P(A\cap D)=\frac{20}{60}=\frac{1}{3}\approx0.33\), \(P(A)\times P(D)=0.6\times0.5 = 0.3
eq\frac{1}{3}\), so \(A\) and \(D\) are not independent.

Option B and D

\(P(B)=\frac{24}{60}=0.4\), \(P(D)=\frac{30}{60}=0.5\), \(P(B\cap D)=\frac{10}{60}=\frac{1}{6}\approx0.17\), \(P(B)\times P(D)=0.4\times0.5 = 0.2
eq\frac{1}{6}\), so \(B\) and \(D\) are not independent.

Option B and E

\(P(B)=\frac{24}{60}=0.4\), \(P(E)=\frac{10}{60}=\frac{1}{6}\approx0.17\), \(P(B\cap E)=\frac{6}{60}=0.1\), \(P(B)\times P(E)=0.4\times\frac{1}{6}=\frac{2}{30}\approx0.07
eq0.1\), so \(B\) and \(E\) are not independent.

Answer:

A. A and C