Probability of A Winning Dept Head Vote w/ 5 Faculty Members

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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?

My answer:

We will say $C$ will be the event that A remains ahead throughout the vote count.

Total number of ways for the three A's to be tallied: ${5 \choose3 } = 10$

In order for A to remain ahead it must be the case that the first two tallies go to A. After that there remain three slips to be tallied: one A and two B. There are ${3 \choose1 } = 3$ ways for the one remaining A to be tallied. One of these ways (BBA) results in A and B having the same number of tallies before the last slip is chosen. Thus,

$P(C) = (3 - 1) / 10 = 2 / 10 = 0.2$

Is this correct?
 
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Hello, KyleM!

I agree with your reasoning and your answer.

To double-check, I listed the [tex]{5\choose3,2}=10[/tex] outcomes.

. . [tex]\begin{array}{ccc} \color{blue}{AAABB} &&ABBAA \\ \color{blue}{AABAB} && BAAAB \\ AABBA && BAABA \\ ABAAB && BABAA \\ ABABA && BBAAA \end{array}[/tex]

Only in the first two does A's votes constantly exceed B's votes.