February 6th, 2011, 03:34 PM #1 Newbie   Joined: Feb 2011 Posts: 3 Thanks: 0 Probability Question 1. After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students (1,2,3, and 4) who claim to have left books. One possible outcome is that 1 receives 2�s book, 2 receives 4�s book, 3 receives his or her book, and 4 receives 1�s book. This outcome can be abbreviated (2,4,3,1). (A) List all outcomes. (B) Assuming equally likely outcomes, what is the probability that exactly two books are returned to their correct owners? (C) What is the probability that exactly one of the four students receives his or her own book? (D) What is the probability that exactly three receive their own book? (E) What is the probability that at least two of the four students receive their own books? I am a little unsure about how to handle this question because there are 4 events was wondering if someone could take me through it. February 6th, 2011, 09:11 PM   #2
Math Team

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Re: Probability Question

Hello, skate_rds!

The problem is straight forward.

Quote:
 1. After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students 1,2,3, and 4) who claim to have left books. One possible outcome is that 1 receives 2�s book, 2 receives 4�s book, 3 receives his or her book, and 4 receives 1�s book. This outcome can be abbreviated (2,4,3,1). (A) List all outcomes.

[color=beige]. . [/color]

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 (B) Assuming equally likely outcomes, what is the probability that exactly two books are returned to their correct owners?

There are six cases.

[color=beige]. . [/color]

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 (C) What is the probability that exactly one of the four students receives his or her own book?

There are eight cases.

[color=beige]. . [/color]

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 (D) What is the probability that exactly three receive their own book?

This is a trick question.
It is impossible for three to receive their book . . . and the fourth doesn't.

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 (E) What is the probability that at least two of the four students receive their own books?

There are six ways for two students to get their own books.
There are no ways for three students to get their own books.
There is one way for all four students to get their own books. February 6th, 2011, 10:12 PM #3 Newbie   Joined: Feb 2011 Posts: 3 Thanks: 0 Re: Probability Question haha i figured it out shortly after posting this you are helping me confirm my answers though your post was greatly appreciated. My confusion was I thought my teacher wanted to show this using a one line formula. March 17th, 2011, 08:37 AM #4 Newbie   Joined: Mar 2011 Posts: 1 Thanks: 0 Re: Probability Question Hey, I need help structurally. Question: What is the probability that an integer between 1 and 9,999 has exactly one 8 and one 9? One answer I came up with was: 4! * 8*8 = 1536 My reasoning: Set {8,9,x,x} where x is any number: 0,1,2,3,4,5,6,7 (total of 8 choices as we don't want to count 8 and 9). a) Given four numbers, there are 4! ways to arrange them b) 8 ways to pick the 3rd number and 8 ways to pick the fourth number. so 1536/(10^4) However, I decided to try it another way and I don't know what case I am failing to count as I get a different answer. a) _ _ = there are 2! = 2 ways to arrange set {(8,9} b)_ _ _ = there are 3!*7 = 42 ways to arrange set {8,9,y} where y is any number: 1,2,3,4,5,6,7 (total of 7 choices as we don't want to count 8,9,0) c) _ _ _ = there are 3!=6 ways to arrange set {8,9,0} . However we don't want to count any sequence where the 0 is the first number so we subtract 2! (from part a)) to get 4 d) _ _ _ _ = there are 4!*7*7 =1176 ways to arrange set {8,9,y,y} e) _ _ _ _ = there are 4!*7 =168 ways to arrange set {8,9,y,0}. However we don't want to count any sequence where the 0 is the first number so we subtract 3!*7 (from part b)) to get 126 f) _ _ _ _ = there are 4! = 24 ways to arrange set {8,9,0,0}. However we don't want to count an sequence beginning with 0 so we subtract 3! (from part c)) to get 18 when we add all these subcases up we get (2)+(42+4)+(1176+126+1 = 1368 1368/(10^4) I don't understand why I am getting a different number can someone explain what I am not taking into account? Tags probability, question ,

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