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)?

Answer 1

The probability that candidate A remains ahead of candidate B throughout the vote count is 1/2.

Step-by-step explanation:

To find the probability that candidate A remains ahead of candidate B throughout the vote count, we need to consider all possible orders in which the slips can be selected for tallying. There are a total of 5 slips, with 3 being votes for A and 2 being votes for B.

Out of the 5 slips, there are 3 positions where the slips for A can be placed. Once the slips for A are placed, the remaining 2 slips for B can be placed in the remaining positions. Therefore, the total number of possible orders is 3 * 2 = 6.

Out of these 6 orders, only 3 orders have A remaining ahead of B throughout the vote count. Therefore, the probability is 3/6 = 1/2.