Question

A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call.
a. In words, define the Random Variable X.


b. List the values that X may take on.


c. Give the distribution of X. X∼


d. On average, how many dealerships would we expect her to have to call until she finds one that has the car?


e. Find the probability that she must call at most 4 dealerships.


f. Find the probability that she must call 3 or 4 dealerships.

Answer 1

Answer:

(a) X = number of dealerships the customer needs to call before she finds a used red Miata car.

(b) X = 1, 2, 3, 4,...

(c) The distribution of X is, $X\sim Geometric\ (p=0.28)$.

(d) The expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.

(e) The probability that she must call at most 4 dealerships is 0.7313.

(f) The probability that she must call 3 or 4 dealerships is 0.2497.

Step-by-step explanation:

(a)

The random variable X can be defined as the number of dealerships the customer needs to call before she finds a used red Miata car.

(b)

The values that the random variable X can assume are:

X = 1, 2, 3, 4,...

(c)

The random variable X is defined as the number of trials before the first success. Each trial is independent of the others. And each trial has a similar probability of success.

This implies that the random variable X follows a Geometric distribution.

The probability of success, i.e. the probability that any independent dealership will have the car is p = 0.28.

Thus, the distribution of X is, $X\sim Geometric\ (p=0.28)$.

(d)

The expected value of a Geometric distribution is:

$E(X)=\frac{1}{p}$

Compute the expected number of dealerships would we expect her to call until she finds one that has the car as follows:

$E(X)=\frac{1}{p}$

         $=\frac{1}{0.28}$

         $=3.57$

Thus, the expected number of dealerships would we expect her to call until she finds one that has the car is 3.57.

(e)

Compute the probability that she must call at most 4 dealerships as follows:

P (X ≤ 4) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

              $=\sum\limits^{4}_{x=1}{(1-0.28)^{x-1}\times 0.28}\\=0.28+0.2016+0.1452+0.1045\\=0.7313$

Thus, the probability that she must call at most 4 dealerships is 0.7313.

(f)

Compute the probability that she must call 3 or 4 dealerships as follows:

P (X = 3 or X = 4) = P (X = 3) + P (X = 4)

                            $=[(1-0.28)^{2-1}\times 0.28]+[(1-0.28)^{4-1}\times 0.28]\\=0.14515+0.10451\\=0.24966\\\approx0.2497$

Thus, the probability that she must call 3 or 4 dealerships is 0.2497.