Question

9- Turns out that the percentages for donuts sold are glazed 22%, maple bars 30%, fritters 19%, old fashioned 10%, filled 11% and other 8%. If you select a dozen at random (weird I know) what is the probability that you get 3 glazed, 4 maple bars, 2 fritters, 1 filled and 2 old fashioned?

Answer 1

Solution

Back-up Theory

If probability of selecting one donut of type A is p, then probability of selecting k donuts of type A = p^k ………………………………………………………………………...............……… (1)

If one dozen of donuts is constituted by n_i donuts of type A_i, where i = 1 to 5 and ∑[1,5]n_i = 12, number of possible combinations = (12!)/{(n₁!)(n₂!)(n₃!)(n₄!)(n₅!)}………...........………..(2)

Now, to work out the solution,

Given data =>

P(one glazed donut) = 0.22

P(one maple bar donut) = 0.3

P(one fritters donut) = 0.19

P(one old fashioned donut) = 0.10

P(one filled donut) = 0.11

P(one other type donut) = 0.08

[Note: total probability is 1]

Vide (1) under Back-up Theory, probability of 3 glazed, 4 maple bars, 2 fritters, 1 filled and 2 old fashioned = 0.22³ x 0.3⁴ x 0.19² x 0.11¹ x 0.1² = P, say ……………………………….(3)

Vide (2) under Back-up Theory, number of possible combinations of 3 glazed, 4 maple bars, 2 fritters, 1 filled and 2 old fashioned = (12!)/{(3!)(4!)(2!)(1!)(2!)} = N, say …………..........….(4)

(3) and (4) => required probability = NP = 0.0028 ANSWER

Details of Excel Calculations

x	n	x^n
0.22	3	0.010648
0.3	4	0.0081
0.19	2	0.0361
0.11	1	0.11
0.1	2	0.01
P = Product		3.4249E-09

n	n!
3	6
4	24
2	2
1	1
2	2
Product P1	576
12	479001600
N = (12!)/P1	831600
NP	0.00284818

DONE