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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 $A$ remain ahead of $B$ throughout the tally?

(a) 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) {AAAABBB, AAABABB, AAABBAB, AABAABB, AABABAB}

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{'transcript': "so let one represent a vote for Candidate A and negative one of vote for Candidate B than a random ization of the 12 A's and eat bees can be achieved by sampling without replacement from a vector of ones and noted ones so 12 ones and eaten it once not to keep track of how far ahead candidate a stands. As each vote is counted, we can use the cume some command or accumulated. So so as long as he's ahead, the cumulative total will be positive. And if any of you ever tied the cumulative, some will be zero and a negative. Accumulative some. And here's the bee has taken the lead. So that's the dis our description, or just of the code here, and one execution of the code returned equals 2013 Soapy hat is equal to a four point"}