There are several scenarios described below. For each of them, do the following (note: R. V. means random variable) (1) Define the R. V.--- that means something like, "Let X be the number of people who….." (2) Define the distribution and parameter(s) of the R. V. (3) Give the support of the R. V. (4) Write the probability statement related to the information being sought.

a) The Ohio Bureau of Motor Vehicles states that 7 out of 8 people pass the written driver’s test.

Let X be the number of test given by the test taker to pass out.

So X~Geometric(p) where p=Probability that for a particular test anyone will pass the written test=7/8

and here support of X be equal to 1,2,3, i.e X is a Natural number

So ,probability that he will pass the written test in fewer than 4 tries

=P(X<4)

=\sum _{x=0}^{3}p(1-p)^{x-1}

b) LAIMO manufacturing company makes parts for the auto industry. Approximately 3% of the parts it makes are defective.

So let X=number of non defective parts sampled before the 3rd defective part is sampled

then X~Negative Binomial(r,p) where here r=3 and p=Probability that a randomly selected part is defective= 0.03

where support of X is {0,1,2,3,...}

So the probability that the third defective part is the 20th one sampled.

P(X=20-3=17)

=\binom{r+16}{17}p^r(1-p)^{16}

c) A BigMart store is going to hire 3 new cashiers. It has 18 applicants (10 male, 8 female) for these 3 cashier jobs.

So let X be number of female cashier appointed.

Here X~Hypergeometric(3,8,18) where

f(x)=P(X=x)

=\left\{\begin{matrix} \frac{\binom{8}{x}.\binom{10}{3-x}}{\binom{18}{3}} & ,x=0,1,2,3\\ 0 & ,otherwise \end{matrix}\right.

So the probability that none of the positions are filled by females

=P(X=0)

d) A gardener is inspecting the fall flowers in her garden. She notices, on average, 4 bugs on a flower. She randomly picks one flower from her garden.

Let X be the numbers of bugs on that flower

So X follows Poisson distribution with mean 4 where support of X is {0,1,2,3}

So the probability that the flower she picked has at least one bug on it

=P(X\geq 1)=1-P(X=0)

=1-e^{-4}\frac{4^x}{x!}|_{x=0}

e) A student is taking a true/false test that consists of 15 questions. Based on past performance the student has approximately a 70% chance of getting any individual question correct.

So let X be the number of questions that are correct among those 15 questions.

so X~Binomial(n,p) where n=15 and p=Probability that he get an individual question correct =0.7

where support of X be {0,1,2,3,...,15}

So the probability that the student gets at least 60% of the questions on the test correct or 15x60%=9 questions are correct

=P(X\geq9)

=\sum _{x=9}^{15}\binom{15}{x}p^x(1-p)^{15-x}

f) A certain radio station’s phone lines are busy approximately 95% of the time when trying to call during a contest.

Let X denotes the number of calls to get into the contest.

So X~Geometric(p) where p=Probability that in a call I get through into the contest=1-0.95=0.05

support of x={1,2,3,}

So the probability that the 4 th time you call is the 1st time you get through during a contest.

=P(X=4)

=p(1-p)^4

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads