Gummi bears come in twelve flavors, and you have one of each flavor. Suppose you

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) (1 point) What is the probability that Adam gets exactly three gummi bears?

(b) (3 points) If you know that each person recieved at least one gummi bear, what is the probability that Adam gets exactly three gummi bears?

(c) (2 points) What is the expected number of gummi bears that Adam gets?

(d) (2 points) If you know that each person recieved at least one gummi bear, what is the expected number of gummi bears that Adam gets?

(e) (2 points) 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

Solution

Given each person is equally likely to be chosen for each gummi bear, chance of Adam getting selected at any draw is 1/3 and the chance he is not selected at any draw is 2/3.

Part (a)

Probability that Adam gets exactly three gummi bears = P(Adam gets selected at 3 draws and not selected at the remaining 9 draws) = (1/3)3(2/3)9 = 29/312.

Now, the 3 draws where Adam gets selected can be any 3 out of 12 draws in 12C3

= 220.

Thus, probability that Adam gets exactly three gummi bears = 220 x (29/312)

= 0.21186 ANSWER

Part (b)

Probability that Adam gets exactly three gummi bears given each person received at least one gummi bear

= P(of the remaining 9 draws after assigning one gummi bear to each one, Adam gets selected at 2 draws and not selected at 7 draws)

9C2(1/3)2(2/3)7 = 0.23411 ANSWER

Part (c)

Let X = Number of gummi bears that Adam gets. Then, X = number of draws out of 12 draws Adam gets selected and X ~ B(12, 1/3). So, expected number of gummi bears that Adam gets = mean of B(12, 1/3) = 12 x (1/3) = 4 ANSWER

Part (d)

Let Y = Number of gummi bears that Adam gets, given each person received at least one gummi bear Then, Y = number of draws Adam gets selected in the remaining 9 draws and Y ~ B(9, 1/3). So, expected number of gummi bears that Adam gets given each person received at least one gummi bear

= 1 + mean of B(9, 1/3) = 4 ANSWER

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