Question

An academic department has just completed voting by secret ballot for a department head. The ballot box contains four slips with votes for candidate A and three slips with votes for candidate B. Suppose these slips are removed from the box one by one.
a. List all possible outcomes.
b. Suppose running tally is kept as slips are removed. For what outcomes does A remain ahead of B throughout the tally?

Answer 1

Answer:

(a) There are 35 possible outcomes.

(b) S = {AABAABB, AABABAB, AAABABB, AAABBAB, AAAABBB}

Step-by-step explanation:

Let's assume that the voting slips are marked as A for candidate A and B for candidate B.

There are 4 slips marked as A and 3 slips marked as B.

(a)

There are a total of 7 slips: 4 of A's and 3 of B's.

The number of arrangements of these slips can be determined by computing the permutation of theses 7 slips.

$Permutation=\frac{n!}{n_{A}!\times n_{B}!} =\frac{7!}{4!\times 3!} =\frac{5040}{24\times6}=35$

Thus, there are 35 possible outcomes when the slips are removed from the box one by one.

The outcomes are

S = {BBBAAAA, BBABAAA, BBAABAA, BBAAABA, BBAAAAB, BABBAAA,  BABABAA, BABAABA, BABAAAB, BAABBAA, BAABABA, BAABAAB, BAAABBA,  BAAABAB, BAAAABB, ABBBAAA, ABBABAA, ABBAABA, ABBAAAB, ABABBAA,  ABABABA, ABABAAB, ABAABBA, ABAABAB, ABAAABB, AABBBAA, AABBABA, AABBAAB, AABABBA, AABABAB, AABAABB, AAABBBA, AAABBAB, AAABABB,  AAAABBB}

(b)

The slips are drawn one by one.

The tallies where A remains ahead of B are such that the count for A is always more than B. This implies that the count of A should always start with 2, 3 or 4.

The possible outcomes are:

S = {AABAABB, AABABAB, AAABABB, AAABBAB, AAAABBB}