91Ó°ÊÓ

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:

$$\begin{gathered} P(X≤4)=P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ =\underset{x=1}{\overset{4}{∑}}{(1-0.28)}^{x-1}×0.28 \\ =0.28+0.2016+0.1452+0.1045 \\ =0.7313 \end{gathered}$$

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

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