Question

An academic department with five faculty members narrowed its choice for department head to either candidate A or candidate B. Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for A and two for B. If the slips are selected for tallying in random order, what is the probability that A remains ahead of B throughout the vote count (e.g., this event occurs if the selected ordering is AABAB, but not for ABBAA)?

Short answer

The probability is \({\rm{0}}{\rm{.2}}{\rm{. }}\) that A remains ahead of B throughout the vote count.

Step-by-step solution

Definition

The term "probability" simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Finding the probability

We can list all outcomes, see what outcomes are favorable and use

\({\rm{P(A) = }}\frac{{{\rm{ \# favorableoutcomesinA}}}}{{{\rm{ \# outcomesinthesamplespace}}}}{\rm{.}}\)

There are

\(\left( {\begin{aligned}{\rm{5}}\\{\rm{3}}\end{aligned}} \right)\)

ways to select positions for A votes (or equally

\(\left( {\begin{aligned}{\rm{5}}\\{\rm{2}}\end{aligned}} \right)\)

ways to select positions for B votes):

\({\rm{BBAAA,BABAA,BAABA,BAAAB,ABBAA,ABABA,ABAAB,AABBA,AABAB,AAABB}}\)

and only two have \({\rm{A}}\)ahead of \({\rm{B}}\) trough out the vote - \({\rm{AABAB}}\) and\({\rm{AAABB}}\).

All outcomes are equally probable therefore we have that the probability that \({\rm{A}}\) remains ahead of \({\rm{B}}\) trough out the vote is

\(\begin{aligned}{\rm{P(A) = }}\frac{{{\rm{\# favorable outcomes in A }}}}{{{\rm{ \# out comes in the sample space}}}}\\{\rm{ = }}\frac{{\rm{2}}}{{{\rm{10}}}}\\{\rm{ = 0}}{\rm{.2}}{\rm{. }}\end{aligned}\)

Therefore, the probability is \({\rm{0}}{\rm{.2}}{\rm{. }}\)