Subject: Random chewing
Date: Wed, 31 Dec 1969 19:00:00 +0000
From: Tarra Williamson

I desperatly need help for this question from my high school Finite course. Any method or suggestions are appreciated! Thanks Everyone!

A cool dude has purchased two chewing gum packs, each containg 10 sticks. Whenever he wants to impress a girl he takes (randomly) one of the pack from his pocket and starts chewing. After a while one of the packs will be empty and the other will contain k sticks. (1 <= k <= 10)

1. What is the probablity that k is 1? 5? 9?
2. What is the most likely value for k?
3. Ann and Betty have made a bet. If K is less than 3 Ann pays Betty \$10. If k is greater than 3 Betty pays Ann \$d. What value for d would make this game fair?

***

A Beginning!

First assume that the packs contain only 3 sticks each. Denote by x the first pack and by y the second.

k=3 for x,x,x or y,y,y
k=2 for x,x,y,x, or x,y,x,x, or y,x,x,x or y,y,x,y, or y,x,y,y or x,y,y,y
k=1 for the rest

Therefore

P(k=3)=2/8=1/4
P(k=2)=6/16=3/8
P(k=1)=1-1/4-3/8=3/8
k=1 and k=2 are equally the most likely

***

Thanks again!

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