Question

In: Statistics and Probability

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.

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 a running tally is kept as slips are removed. For what outcomes does Aremain ahead of B throughout the tally?

Solutions

Expert Solution

Solution

• Step1

Denote with A when we remove a slip with candidate A's votes and with B when we remove a slip with candidate B's votes.

(a): We have four slips with votes for A and three slips with votes for B, therefore we have four A's and three B's from which we need to create the outcomes. Every outcome should be contained of 4+3 = 7 spaces, for example AAAABBB (which would mean that first four slips with votes removed were from the candidate A, and the next three slips with votes removed were from the candidate B). All possible outcomes are

S ={BBBAAAA, BBABAAA, BBAABAA,BBAAABA, BBAAAAB,BABBAAA

BAAAABB, BABABAA, BABAABA,BABAAAB,BAABBAA, BAABABA,

BAABAAB, BAAABBA, BAAABAB,ABBBAAA,ABBABAA, ABBAABA,

ABBAAAB, ABABBAA, ABABABA,ABABAAB,ABAABBA, ABAABAB,

ABAAABB, AABBBAA, AABBABA,AABBAAB,AABABBA,AABABAB

AABAABB, AAABBBA, AAABBAB,AAABABB,AAAABBB}.

Hopefully, we didn't miss any event There are some methods for listing all the possible events of this type-start with the same letters one next to another and move one by one until you rearrange them up to the other way you started (this is not a known method, depending on number of letters it gets harder).

• Step2

(b): We need all events where number of A's in the x_1x_2 . . . xi are always bigger than number of B's for every i = 1, 2,...,7 where x_1x_2 . . . xi are first i spaces in particular outcome. For example, if the outcome is AAABABB, for i = 4 we have AAAB and here number of B's is one and number of A's is 3, 3 > 1, when this is true for every i = 1,2,...,7 we can conclude that the particular outcome belongs to our event Therefore, outcomes for which this is true are

⇒ {AAAABBB, AAABABB, AAABBAB,AABAABB,AABABAB} .


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