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)?

The probability that the first two votes drawn are both for candidate a is given by:
3C2/5C2 = 3/10
Having drawn two votes for candidate a on the first two draws, there are 2 votes for candidate b and one vote for candidate a remaining. The probability that a vote for candidate b will be drawn on the third draw is:
2/3.
After the first three draws, there reains one vote for candidate a and one vote for candidate b. The probability that a vote for candidate a will be drawn on the fourth draw is:
1/2.
The probability of the ordering aabab is therefore given by:

The answer is: 0.1.

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.