An employee of the records office at a certain university

Chapter 3, Problem 92E

(choose chapter or problem)

Get Unlimited 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?

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.

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back