Mind Boggled : Not exactly a fixed number of Trials, but Independent Events?

Question

And so, I have a math question for my stats homework that goes like this :

A car dealer has a list of 15 cars. The probability of selling one car during a typical week is 40%. The chance of selling any one car is independent of selling another. Find the probability of selling more than 4 cars in one week.

I've been having a headache trying to solve this - is it even possible for the probability of one car to be sold to remain independent throughout, especially when the number of cars left in the set of 15 will gradually diminish as each subsequent one is sold?

I have also tried using the Binomial distribution method to solve this, and got a result of 0.78 probability of P(X=5 ≤ 15) which seems highly unlikely.

Set of 15(no. of trials?) p=0.4 (?) X(random variable) = 5 ≤ 15 (since more than 4 in this case, means selling at least 5, and at most 15)

Also on second thought, i do not think this qualifies as binomial, unless each car being sold will be replaced and resulting in a purely fixed number of 15 cars.

Anyone here has any idea how to work on this?

Answer 1

One difficulty is to understand what is meant by "[T]he probability of selling one car during a typical week is 40%."

(a) Does that mean the probability is $0.4$ that any one car is sold?

(b) Does it mean that the probability is $0.4$ that exactly one car is sold in a week?

(c) Does it mean the probability is $0.4$ that at least one car is sold in a week?

(d) Or, does it mean the average number of cars sold per week is $0.4,$ and then find the probability sales in a week exceed 4?

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This is an interesting problem in building probability models for real life situations. I'll show several (speculative) possible ways to approach the problem.

One approach is to choose interpretation (c) and let the number $X$ of cars sold in a week have a Poisson distribution with appropriate rate $\lambda.$ $$P(X \ge 1) = 1 - P(X = 0) = 1 - e^{-\lambda} = 0.4,$$ so that $\lambda = 3.2189.$

This is an approximate model with individual car sales modeled as random events. It is easy to use R to find the distribution of $X.$ [Ignore numbers in brackets.]

x = 0:15;  pdf = round(dpois(x, 3.2189), 4)
cbind(x, pdf)
       x    pdf
 [1,]  0 0.0400
 [2,]  1 0.1288
 [3,]  2 0.2072
 [4,]  3 0.2223
 [5,]  4 0.1789
 [6,]  5 0.1152
 [7,]  6 0.0618
 [8,]  7 0.0284
 [9,]  8 0.0114
[10,]  9 0.0041
[11,] 10 0.0013
[12,] 11 0.0004
[13,] 12 0.0001
[14,] 13 0.0000
[15,] 14 0.0000
[16,] 15 0.0000

Technically, there is no limit on how many cars might be sold, but realistically there is almost no chance of selling more than 13. Also, $P(X > 4) = 0.2228.$

1 - ppois(4, 3.2189)
[1] 0.2227613

Somewhat similar methods can be used to make binomial and Poisson models for alternative interpretations, solving for either Poisson $\lambda$ or binomial $p.$ [I share your misgivings about binomial models, but I think any model will have to make interpretations and assumptions, maybe binomial models are OK as approximations.]

For the model $\mathsf{Binom}(n = 15, p=.4),$ the distribution is shown below:

x = 0:15;  pdf = dbinom(x, 15, .4)
cbind(x, round(pdf,4))
       x       
 [1,]  0 0.0005
 [2,]  1 0.0047
 [3,]  2 0.0219
 [4,]  3 0.0634
 [5,]  4 0.1268
 [6,]  5 0.1859
 [7,]  6 0.2066
 [8,]  7 0.1771
 [9,]  8 0.1181
[10,]  9 0.0612
[11,] 10 0.0245
[12,] 11 0.0074
[13,] 12 0.0016
[14,] 13 0.0003
[15,] 14 0.0000
[16,] 15 0.0000

As several have said, this leads to $P(X > 4) = 0.7827.$

1 - pbinom(4, 15, .4)
[1] 0.7827223

Using interpretation (d) and $X \sim \mathsf{Pois}(\lambda = 0.4),$ one obtains $P(X > 4) = .000061.$ Most of the probability is for lower sales.

1 - ppois(4, .4)
[1] 6.124333e-05

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If your professor intended this problem to provoke you into thinking about thorny issues in making realistic probability models for real life situations, then I guess the problem is a success.

Answer 2

my professor has just got back to me - and yes, it is indeed binomial!

The Poisson method looks really interesting though. Thanks everyone for all the insights!