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An employee of the records office at a certain university
Chapter 3, Problem 92E(choose chapter or problem)
An employee of the records office at a certain university currently has ten forms on his desk awaiting processing. Six of these are withdrawal petitions and the other four are course substitution requests. a. ?If he randomly selects six of these forms to give to a subordinate, what is the probability that only one of the two types of forms remains on his desk? b. ?Suppose he has time to process only four of these forms before leaving for the day. If these four are randomly selected one by one, what is the probability that each succeeding form is of a different type from its predecessor?
Questions & Answers
QUESTION:
An employee of the records office at a certain university currently has ten forms on his desk awaiting processing. Six of these are withdrawal petitions and the other four are course substitution requests. a. ?If he randomly selects six of these forms to give to a subordinate, what is the probability that only one of the two types of forms remains on his desk? b. ?Suppose he has time to process only four of these forms before leaving for the day. If these four are randomly selected one by one, what is the probability that each succeeding form is of a different type from its predecessor?
ANSWER:Answer : Step 1 of 2 : Given, An employee of the records office at a certain university currently has ten forms on his desk awaiting processing. Six of these are withdrawal petitions and the other four are course substitution requests. a) If he randomly selects six of these forms to give to a subordinate, the claim is to find the probability that only one of the two types of forms remains on his desk W = withdrawal positions = 6 S = Course substitution request = 4 Either W of the forms are selected or all of the course substitution forms are selected and 2 of the W are selected. The total number of ways this can be done is 6C6 +6C2 n! C(n, r) = r!(n r)! C(6, 6) =1 6! C(6, 2) = 2!(6 2)! = 15 The total number of ways this can be done is 15 The total number of ways 6 can be chosen is 10C6 10! C(10, 6) = 6!(10 6)! = 210 The required probability is 16/210 = 0.07619.