Question

Consider a political discussion group consisting of 8 ​Democrats, 5 ​Republicans, and 5 Independents. Suppose that two group members are randomly​ selected, in​ succession, to attend a political convention. Find the probability of selecting two Independents.

Answer 1

Answer:

The probability of selecting two Independents is $\frac{10}{153}$.

Step-by-step explanation:

From the given information it is clear that:

​Democrats = 8

Republicans = 5

Independents = 5

Total number of member in the group is

$8+5+5=18$

We need to find the probability of selecting two Independents.

According to binomial distribution the total number of ways to select r items form n items is

$^{n}C_r=\frac{n!}{r!(n-r)!}$

Total number of ways to select 2 members from 18 members is

$\text{Total possible outcomes}=^{18}C_2=\frac{18!}{2!(18-2)!}=153$

Total number of ways to select 2 members from 5 Independents is

$\text{Favorable outcomes}=^{5}C_2=\frac{5!}{2!(5-2)!}=10$

The probability of selecting two Independents is

$p=\frac{\text{Favorable outcomes}}{\text{Total possible outcomes}}$

$p=\frac{10}{153}$

Therefore the probability of selecting two Independents is $\frac{10}{153}$.