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 <em>A </em>for candidate A and <em>B</em> for candidate B.
There are 4 slips marked as <em>A </em>and 3 slips marked as <em>B</em>.
(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](https://tex.z-dn.net/?f=Permutation%3D%5Cfrac%7Bn%21%7D%7Bn_%7BA%7D%21%5Ctimes%20n_%7BB%7D%21%7D%20%3D%5Cfrac%7B7%21%7D%7B4%21%5Ctimes%203%21%7D%20%3D%5Cfrac%7B5040%7D%7B24%5Ctimes6%7D%3D35)
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 <em>A</em> remains ahead of <em>B</em> are such that the count for <em>A</em> is always more than <em>B</em>. This implies that the count of <em>A</em> should always start with 2, 3 or 4.
The possible outcomes are:
S = {AABAABB, AABABAB, AAABABB, AAABBAB, AAAABBB}