Question

Two candidates face each other in an election. The Democratic candidate is supported by 46% of the population, and the Republican candidate is supported by 54%. In other words, if you randomly chose a voter and asked them who they plan to vote for, there would be a 46% chance they would say they support the Democratic candidate. Suppose you run a poll of 8 people (randomly choose 8 people). What is the probability that less than half of them (3 or fewer) would support the Republican candidate?

Answer 1

Answer:

27.98% probability that less than half of them (3 or fewer) would support the Republican candidate

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they support the Republican candidate, or they do not. The people are chosen at random, which means that the probability of them supporting the republican candidate is independent from other people. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

The Republican candidate is supported by 54%. This means that $p = 0.54$

Suppose you run a poll of 8 people (randomly choose 8 people). What is the probability that less than half of them (3 or fewer) would support the Republican candidate?

This is $P(X \leq 3)$ when $n = 8$.

So

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{8,0}.(0.54)^{0}.(0.46)^{8} = 0.0020$

$P(X = 1) = C_{8,1}.(0.54)^{1}.(0.46)^{7} = 0.0188$

$P(X = 2) = C_{8,2}.(0.54)^{2}.(0.46)^{6} = 0.0774$

$P(X = 3) = C_{8,3}.(0.54)^{3}.(0.46)^{5} = 0.1816$

So

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0020 + 0.0188 + 0.0774 + 0.1816 = 0.2798$

27.98% probability that less than half of them (3 or fewer) would support the Republican candidate