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 four dealerships.

f. Find the probability that she must call three or four dealerships

In a Bernoulli trial, the likelihood of the amount of successive failures before a hit is obtained is represented by a geometrical distribution, which may be a form of discrete probability distribution. A Bernoulli trial may be a test which will only have one among two outcomes: success or failure.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, during this case $X$is that the number of dealership the customer has to call before she finds a second hand red Miata car.

Make the list of values that you just want to use $X$may tackle.

As we are able to see there's an boundary for true at hand so,

localid="1649157268204" $X=1,2,3,4,..........$

The variable $X$refers to the quantity of trials before the primary success. Each trial is independent of others and has similar probability of success.

This implies that variate$X$follows Geometric Distribution.

The probability that any independent dealership will have car is $p=0.28$

Thus, the distribution of X is$X~G(0.28)$

The arithmetic mean of geometric distribution is:

$E\left(X\right)=\frac{1}{p}$, where $p=0.28$

Therefore,

$$\begin{gathered} E\left(X\right)=\frac{1}{p} \\ E\left(X\right)=\frac{1}{0.28} \\ E\left(X\right)=3.57 \end{gathered}$$

The probability that she must call at most four dealerships is:

$$\begin{gathered} P(X\le 4)=P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ =\sum _{x=1}^4{(1-0.28)}^{x-1}\times 0.28 \\ =0.28+0.2016+0.1452+0.1045 \\ =0.7313 \end{gathered}$$

The probability that she must call three or four dealerships is:

$$\begin{gathered} P(X=3orX=4)=P(X=3)+P(X=4) \\ {=\left[\right(1-0.28)}^{3-1}\times (0.28)]+[{(1-0.28)}^{4-1}\times (0.28)] \\ =0.14515+0.10451 \\ =0.2497 \end{gathered}$$