Gummi bears come in twelve flavors, and you have one of each flavor. Suppose you
ID: 3844984 • Letter: G
Question
Gummi bears come in twelve flavors, and you have one of each flavor. Suppose you split your gummi bears among three people (Adam, Beth, Charlie) by randomly selecting a person to receive each gummi bear. Each person is equally likely to be chosen for each gummi bear. (a) What is the probability that Adam gets exactly three gummi bears? (b) If you know that each person received at least one gummi bear, what is the probability that Adam gets exactly three gummi bears? (c) What is the expected number of gummi bears that Adam gets? (d) If you know that each person received at least one gummi bear, what is the expected number of gummi bears that Adam gets? (e) Is there more variance in the number of gummi bears that Adam gets with or without knowing that everyone gets at least one gummi bear? Explain your answer without calculations.Explanation / Answer
(a)
Bears belong to Adam
*
Bears belong to Beth
*
Bears belong to Charlie
BBB
*
Bears belong to Beth
*
Bears belong to Charlie
The probability that Adam gets exactly three gummi bears, P(A=3)= 10/91
(b)
Also distribute 3 gummi bears to Adam
BBB
*
B
*
B
8 positions
If already each person gets at least one gummi bear, the probability that Adam gets exactly three gummi bears= 8/91
(c)
X
Probability to get X by Adam
0
Since C(13,12)=13, P(X=0)=13/91
1
Since C(12,11)=12, P(X=1)=12/91
2
P(X=2)=11/91
3
P(X=3)=10/91
4
P(X=4)=9/91
5
P(X=5)=8/91
6
P(X=6)=7/91
7
P(X=7)=6/91
8
P(X=8)=5/91
9
P(X=9)=4/91
10
P(X=10)=3/91
11
P(X=11)=2/91
12
P(X=12)=1/91
7*(6/91) +8*(5/91)+9*(4/91)+10*(3/91)+11*(2/91)+12*(1/91)
=0+0.132+0.242+0.33+0.396+0.44+0.462+0.462+0.44+0.396+0.33+0.242+0.132
=4.004
Therefore the expected number of gummi bears that Adam gets is 4.004.
Bears belong to Adam
*
Bears belong to Beth
*
Bears belong to Charlie
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