In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once.

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. How many pages do you expect to advertise footwear on them?

e. Is it probable that all twenty will advertise footwear on them? Why or why not?

f. What is the probability that fewer than ten will advertise footwear on them?

g. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution.

h. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it?

i. How many pages do you expect to need to survey in order to find one that advertises footwear?

In a Bernoulli trial, the likelihood of the number of successive failures before a success is obtained is represented by a geometric distribution, which is a sort of discrete probability distribution. A Bernoulli trial is a test that can only have one of two outcomes: success or failure.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of pages that advertise footwear.

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand so,

$X=0,1,2,...........,20.$

The random variable X follows Binomial Distribution which is given by$X~B(n,p)$where$n=20$the number of trials and$p=\frac{29}{192}$is the probability of success.

Thus, the distribution of X is $X~B(20,\frac{29}{192})$

The number of pages to advertise footwear on them is shown through binomial distribution.

$\mu =np$where $\mu$is the number of pages we expect to advertise footwear on them , $n=20$is the number of trials and $p=\frac{29}{192}$is possible success.

$$\begin{gathered} \mu =np \\ \mu =20\times \frac{29}{192} \\ \mu =3.02 \end{gathered}$$

No, there is no probability that all twenty will advertise footwear on them because the probability is zero.

The probability that fewer than ten will advertise footwear on them is as follow:

$P(X\ge 10)=\left(\begin{array}{c}20!\\ 10!10!\end{array}\right){\left(\frac{29}{192}\right)}^{10}{\left(\frac{163}{192}\right)}^{10}=0.9997$

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case X is the number of pages we must survey until we find that one advertises footwear. The distribution is$X~G\left(\frac{29}{192}\right)$

The probability that we only need to survey at most three pages in order to find one that advertises footwear on it is as follow:

$$\begin{gathered} P(X\le 3)=P(X=1)+P(X=2)+P(X=3) \\ =\sum _{x=1}^3{(1-\frac{29}{192})}^{x-1}\times \frac{29}{192} \\ =0.3881 \end{gathered}$$

The number of pages that we expect to survey in order to find one that advertises footwear is calculated using geometric distribution.

$E\left(X\right)=\frac{1}{p}$where $p=\frac{29}{192}$

Thus,

$$\begin{gathered} E\left(X\right)=\frac{1}{p} \\ E\left(X\right)=\frac{1}{{\displaystyle \frac{29}{192}}} \\ E\left(X\right)=6.6207 \end{gathered}$$