Question

Suppose for your company, the probability an individual customer purchases something in-store is 0.74. The probability he or she purchases something online is 0.81. We have described a set of Bernoulli trials with this scenario:

            (a) There are two outcomes – purchase (a “success”) or no purchase,

            (b) The probability of a success is constant across trials, and

(c) Trials are independent of each other, assuming customers are not related or shopping together.

(1) What is the expected number of customers until the first one purchases something in-store?

(2) What is the expected number of customers until the first one purchases something online?

(3) What is the probability that the third customer is the first to purchase something in-store?

(4) What is the probability that the fifth customer is the first to purchase something online?

(5) What is the probability of exactly 7 of the next 12 customers purchasing something in-store?

(6) What is the probability that exactly 7 of the next 12 customers will purchase something online?

(7) What is the probability that no more than 6 of the next 9 customers will purchase something in-store?

(8) What is the probability that at least 8 of the next 11 customers will purchase something online?

Now we want to look at purchasing a specific item. Suppose the probability customers purchase that item in-store is 0.60, while the probability they purchase it online is 0.79.

(9) What is the probability exactly 4 out of the next 7 in-store customers will purchase that item?

(10) What is the probability at least 5 of the next 8 online customers will purchase that item?

(11) What is the probability the 4^th in-store customer will be the first to purchase that item?

(12) What is the probability no more than 5 of the next 11 online customers will purchase that item?

(13) What is the probability that at least 8 of the next 12 in-store customers will purchase that item?

(14) What is the probability the first online customer to purchase that item will be before the 5^th?

Suppose for your store, the mean number of customers at any time of the day is 4.8 (this would be λ). Determine the following probabilities. NOTE these are not based on Bernoulli trials.

(15) What is the probability of having no customers in the store at some point during the day?

(16) What is the probability of having at least 6 customers in the store at any given point?

(17) What is the probability of having no more than three customers in the store at any given point?

(18) What is the probability of having exactly 5 customers in the store at any given point?

(19) If I wanted to know the number of customers before the first one bought something, which model would I use?

            GEOMETRIC            BINOMIAL                POISSON            (circle one)

(20) If I wanted to know how many customers out of a certain number will buy something, which model would I use?

            GEOMETRIC            BINOMIAL                POISSON            (circle one)

Answer 1

Solution Q1 to Q8

Back-up Theory

Geometric distribution is the probability distribution of the number X of Bernoulli trials needed

to get one success. It gives the probability that the first occurrence of success requires k independent trials, each with success probability p. If the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is:

pmf = P(X = k) = (1 - p)^{k – 1}.p, for k = 1, 2, 3, ………................................................................................................................ (1)

Mean: 1/p ..................................................................................................................................................................................(2)

If an experiment is dichotomous, i.e., has only two possible outcomes, say success and failure,

which are mutually exclusive and collectively exhaustive, n repetitions of the experiment are identical, outcomes are independent and probability of a success, say p, is constant, then the random variable X, that represents the number of successes of the experiment, is a binomial variable and it is represented as: X ~ B(n, p) .…………............................…….. (3)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where

n = number of trials and p = probability of one success, then, probability mass function (pmf)

of X is given by p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x},...............................................................................................………..(4)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST]............................................ (4a)

Now to work out the solution,

For Q1 to Q4,

Let

X1 = number of customers until the first one purchases something in-store.......................(5a)

X2 = number of customers until the first one purchases something online. .......................(5b)

Y1 = number of customers out of n customers who purchase something in-store..............(6a)

Y2 = number of customers out of n customers who purchase something online................(6b)

Given,

‘the probability an individual customer purchases something in-store is 0.74’ .................. (7a)

and the probability he or she purchases something online is 0.81.’................................... (7b)

Vide (1), (5a) and (7a), X1 ~ Geometric (0.74) .................................................................. (8a)

Vide (1), (5b) and (7b), X2 ~ Geometric (0.81) .................................................................. (8b)

Vide (3), (6a) and (7a), Y1 ~ B(n, 0.74) ............................................................................. (9a)

Vide (3), (6b) and (7b), Y2 ~ B(n, 0.81) ............................................................................. (9b)

Q1

Vide (5a), (7a) and (2),

Expected number of customers until the first one purchases something in-store

= 1/0.74

= 1.35 Answer 1

Q2

Vide (5b), (7b) and (2),

Expected number of customers until the first one purchases something online

= 1/0.81

= 1.23

Answer 2

Q3

Vide (1), (5a) and (7a),

Probability that the third customer is the first to purchase something in-store

= P(X1 = 3)

= (1 – 0.74)²

= 0.0676 Answer 3

Q4

Vide (1), (5b) and (7b),

Probability that the fifth customer is the first to purchase something online

= P(X2 = 5)

= (1 – 0.81)⁴

= 0.0013 Answer 4

Q5

Vide (9a), (4) and (4a),

Probability of exactly 7 of the next 12 customers purchasing something in-store

P(Y1 = 7/n = 12)

= 0.1143 Answer 5

Q6

Vide (9b), (4) and (4a),

Probability that exactly 7 of the next 12 customers will purchase something online

P(Y2 = 7/n = 12)

= 0.0449 Answer 6

Q7

Vide (9a), (4) and (4a),

Probability that no more than 6 of the next 9 customers will purchase something in-store

P(Y1 ≤ 6/n = 9)

= 0.3713 Answer 7

Q8

Vide (9b), (4) and (4a),

Probability that at least 8 of the next 11 customers will purchase something online?

P(Y2 ≥ 8/n = 11)

= 0.8603 Answer 8

DONE