Answer:
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
= 1008