Question

1. Archway Adventures has compiled data for 2017 on customer orders by region and product group as follows:

	North America	Europe	Rest of World	Total
Children’s Toys	5,400	2,700	900	9,000
Games	4,200	2,700	1,100	8,000
Other	1,300	900	800	3,000
Total	10,900	6,300	2,800	20,000

Let us assume that these are representative of the pattern of orders that they anticipate seeing in 2018.

1. Are the events “Toy” and “Rest of the World” independent? Explain.
2. The rows of the table are mutually exclusive categories. How would our summary be complicated if some orders were for multiple products?

2. 54.5% of Archway’s orders come from North America, 31.5% from Europe and the remaining 14% from the rest of the world. Among North American customers, 3% complained about late or lost deliveries. Similarly, complaints were received from 8% of European customers and 15% of customers from the rest of the world.

1. Construct a probability tree that shows all combinations of where orders originated and whether customers complained. Label all branches in terms of what the event is on the branch and the corresponding probability (e.g., P(A), PA’), P(B|A), P(B’|A),…). Also determine the probability for each ending joint event (e.g., P(Europe and Complain)).
2. Summarize all of the joint probabilities in a probability table.
3. What percentage of Archway customers complained about delivery problems?
4. Among those who complained, what percentage were from Europe?

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.

If A and B are independent, P(A ∩ B) = P(A) x P(B) ..…………............................................………………(2)

Now to work out the solution,

Q1

Part (a)

Let A and B respectively represent the events “Toy” and “Rest of the World”.

Out of a total of 20000 orders, 9000 were for Toys, 2800 came from Rest of the World

and 900 were for Toys from Rest of the World. So, vide (1),

P(A) = 9000/ 20000 = 0.45

P(B) = 2800/200000 = 0.14

P(A ∩ B) = 900/20000 = 0.045

Clearly, P(A ∩ B) is not equal to P(A) x P(B). Hence, vide (2),

Events “Toy” and “Rest of the World” are NOT independent. Answer 1

Part (b)

When some orders are for multiple products, the data presentation would take the following form:

Product(s)	North America	Europe	Rest of World	Total
Children’s Toys only
Games only
Other only
Children’s Toys and Games only
Children’s Toys and Other only
Games and Other only
Total

Answer 2

Q2

The given data in a concise from is:

Details	North America	Europe	Rest of World	Total
Percentage composition	54.5	31.5	14.0	100
Complaint percentage	3	8	15

Part (a)

Let events A, B, C represent the events that orders originated from North America, Europe and Rest of World respectively and D represent the event of a complaint. Then,

P(A) = 0.545

P(B) = 0.315

P(C) = 0.14

P(D/A) = 0.03

P(D/B) = 0.08

P(D/C) = 0.15

And finally,

P(D) = {P(D/A) x P(A)} + {P(D/B) x P(B)} +{P(D/C) x P(C)}

= 0.01635 + 0.02520 + 0.0210

= 0.06255

Answer 3

Part (b)

Summary all of the joint probabilities

Joint Event	Probability
A and D	0.01635
B and D	0.0252
C and D	0.0210

Answer 4

Part (c)

From last part of Answer 3, P(D) = 0.06255. Hence,

percentage of Archway customers who complained about delivery problems

= 100 x 0.06255

= 6.255% Answer 5

Part (d)

Out of 6.255% complaints, 2.52% were from Europe. Hence, among those who complained, percentage that were from Europe

= 100 x (2.52/6.255)

= 40.3% Answer 6

DONE