Question

A cooler has 6 Gatorades, 2 colas, and 4 waters. You select 3 beverages from the cooler at random. Let B denote the number of Gatorade selected and let C denote the number of colas selected. For example, if you grabbed a cola and two waters, then C = 1 and B = 0.

a) construct a joint probability distribution for B and C.

b) compute E[3B-C^2].

Answer 1

Solution

Back-up Theory

Probability of an event E, denoted by P(E) = n/N ………………………………................……….……(1)

where

n = n(E) = Number of outcomes/cases/possibilities favourable to the event E and

N = n(S) = Total number all possible outcomes/cases/possibilities.

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}….................…(2)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN….................…. (2a)

If a discrete random variable, X, has probability function) p(x), x = x₁, x₂, …., x_n, then

Expected value of X = E(X) = Σ{x.p(x)} summed over all possible values of x…….......................…. (3)

Expected value of a function f(x) of variable X = E{f(X)} = Σf(x).p(x)} summed over all possible

values of x………………………………………………………………………....................………………(3a)

Now to work out the solution,

Part (a)

We have 12 beverages, 6 Gatorades, 2 colas, and 4 waters and we select 3 beverages

from the cooler at random.

So, vide (1) and (2), N = ¹²C₃ = 220 .................................................................................................. (4)

C = c and B = b => number of colas = c, number of Gatrode = b and hence number of

waters = 3 – c – b.

Thus, for C = c and B = b, n = (²C_c) x (⁶C_b) x (⁴C_3-c-b) [vide (1), (2)................................................... (5)

And hence joint probability for C = c and B = b is: (5)/220 ....................................................................(6)

For possible values of c and b, joint probability is tabulated below:

c	b	w	n	Joint probability
0	0	3	4	0.0182
0	1	2	36	0.1636
0	2	1	60	0.2727
0	3	0	20	0.0909
1	0	2	12	0.0545
1	1	1	48	0.2182
1	2	0	30	0.1364
2	0	1	4	0.0182
2	1	0	6	0.0273
Total			220	1

Answer 1

Part (b)

E[3B-C²] = Σ{(3b – c²) (p), summed over all possible values of b and c, where p is the corresponding joint probability [vide (3a)]

= 2.9545 Answer 2

Details

c	b	w	prob	3b - c2	prob x (3b - c2)
0	0	3	0.0182	-9	-0.1636
0	1	2	0.1636	-1	-0.1636
0	2	1	0.2727	5	1.3636
0	3	0	0.0909	9	0.8182
1	0	2	0.0545	-4	-0.2182
1	1	1	0.2182	2	0.4364
1	2	0	0.1364	6	0.8182
2	0	1	0.0182	-1	-0.0182
2	1	0	0.0273	3	0.0818
			1		2.9545

DONE