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 recieved 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 recieved 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.

Answer 1

Step-by-step explanation:

From the given gummi bear, the chance that Adam is selected for any draw is 1/3 as well as the chance he is not selected at any draw is 2/3.

a). The probability of Adam getting exactly three gummi bears = P(Adam gets selected at 3 draws and not selected at the remaining 9 draws)

= $(\frac{1}{3})^3 (\frac{2}{3})^9 = \frac{2^9}{3^{12}}$

Now, the 3 draws where Adam gets selected can be any 3 out of 12 draws in ${\overset{12}C}_3$ = 220

Thus, probability of Adam getting three gummi bears = $220 \times \frac{2^9}{3^{12}}$

                                                                                                = 0.21186

b). Probability that Adam will get the three gummi bears given each person will received at the most 1 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) = ${\overset{9}C}_2 (\frac{1}{3})^2 (\frac{2}{3})^7$

                                                                                               = 0.23411

c). Let X = Number of the gummi bears which Adam will get. Then, X = number of draws out of 12 draws Adam gets selected and X ~ B(12, 1/3). So,  Adam will get gummi bears= mean of B(12, 1/3) = 12 x (1/3) = 4

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

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

e). When every one gets atleast one gummi bear, the new sample size will be 9 and so we can say that there is a reduction in variance.