Probability Question: Order of Sampling versus when order does not matter?

Question

50 Percent of voters support the president. What is the probability that, in a sample taken, 3 support the president and 2 do not? How does this differ from the probability found if the first three support the president and the last two do not?

Answer 1

Let S represent supports and A represent is against. In the first case, you count the case where you get SASAS and a number of other orders. How many are there? In the second case you only count SSSAA, so it will be a factor less likely.

Answer 2

I suppose the following method will appear in your course soon, if it hasn't already. Let $X$ be the number of Supporters out of five randomly chosen voters. Then $X$ has a binomial random variable with $n = 5$ and $p = P(\text{Support}) = 1/2.$ This is sometimes written $X \sim \mathsf{Binom}(5, 1/2).$

Then the answer to the first question is $$P(X = 3) = {5 \choose 3}p^3(1-p)^2 = 10\left(\frac 1 2\right)^5 = 5/16.$$ The role of the factor ${5 \choose 3} = 10$ is to account for the ten arrangements SSSAA, SSASA, ..., AASSS.

To answer the second question, you need to account for only one of the ten arrangements.