Question

Children are frequently excited to buy breakfast cereal in an effort to “collect all six action figures.” Assume that there are six action figures and each cereal box contains exactly one of the six with each figure being equally likely to be in a box. How many boxes do parents typically have to buy to satisfy their child’s appetite? Using the Random Numbers app, select “1” as the minimum value and “6” as the maximum and generate 50 numbers. Count how many generated numbers are required so that each of the integers, 1 through 6, appears at least once. Repeat this experiment (simulation) several times to get a rough idea of how many generated integers (representing the cereal boxes purchased) are required from each simulation to obtain all six action figures.

Questions for discussion:

1. After several simulations (>25), what would be the estimated probability of collecting all six action figures in 20 purchases?

2. How could you estimate the likelihood of collecting all six action figures in 10 or fewer purchases? What would your estimate be?

Answer 1

Random Integer Generator Here are your random numbers: 4 4 4 4 4 4 4 4 4 4

So, 12 generated numbers are required so that each of the integers, 1 through 6, appears at least once.

Let us run several simulations of 50 integers where minimum is 1 and maximum is 6.

Simulation number	Required numbers to be generated
1	12
2	8
3	12
4	25
5	7
6	10
7	19
8	7
9	10
10	17
11	21
12	17
13	13
14	12
15	34
16	38
17	17
18	18
19	9
20	8
21	12
22	21
23	14
24	19
25	11
26	14
27	11
28	11
29	14
30	15

Thus, from 30 simulations above, we got a minimum required numbers to be generated is 7 and maximum is 38 in order to obtain all 6 action figures at least once. 12 occurred most of the times (4 times).

Mean or expected numbers to be generated i.e., expected number of purchases =(12+8+12+.....14+15)/30 =15.2 which is rounded to 15 (because number of purchases cannot be a fraction).

1.

To find probability, I shall take 38 as the total required purchases because it is the maximum number of purchases required.

To find out the probability, P of collecting all six action figures is in 20 purchases, we need to see how many of those numbers that are less than or equal to 20 in the above table and divide it by 30. Hence,

P= 25/30 =0.833

2.

The probability of collecting all six action figures in 10 or fewer purchases is, P =7/30 =0.233

The likelihood of collecting all six action figures in 10 or fewer purchases is, L(0.233|7) =30C₇ (0.233)⁷ (1-0.233)^30-7 =0.170