Consider exercise 4.45 on page 255 of your textbook.

Suppose that a student is taking a multiple-choice exam in which each question
has four choices. Assuming that she has no knowledge of the correct to any of
the questions, she has decided on a strategy in which she will place four balls
(marked A, B, C, and D) into a box. She randomly selects one ball for each
question and replaces the ball in the box. The marking on the ball will
determine her answer to the question.

Using this method (not recommended for MA 216!), the probability of getting
a correct answer would be 1/4 = .25 for each question. This is a binomial
situation with n=5 and p=.25.

Note that all of the properties of the binomial distribution are satisfied:

- 1.
- The student is sampling with replacement.
- 2.
- There are only two outcomes: A correct answer (Success) or an incorrect answer (Failure). We are not concerned with which wrong answer it might be among the three possibilities.
- 3.
- p=.25 is constant from draw to draw.
- 4.
- The draws are independent- each one does not effect the next.

This page will contain the following solutions:

**First, let's solve this problem by hand, using the binomial formula
(page 246).**

(a)(1) If there are five multiple-choice questions on the exam, what
is the probability that she will get five questions correct?

(a)(2) What is the probability that she will get at least four
questions correct?

.01465 + .00098 = .01563

(a)(3) What is the probability that she will get no questions correct?

(a)(4) What is the probability that she will get no more than two
questions correct?

.23730 + .39551 + .26367 = .89648

(b) The assumptions of the binomial that were previously mentioned.

(c) What are the average and the standard deviation of the number of
questions that she will get correct in (a)?

(d) This cannot be done very easily by hand...

**Now, let's go back and solve part (a) using the TI-83.**

Assuming the lists *L*_{1}, *L*_{2}, and *L*_{3} are empty, put the sample
space {0, 1, 2, 3, 4, 5} in *L*_{1}, using the `seq`

function under
`LIST OPS`

. Simply insert "`A`

" for the expression and variable,
since we just want the sequence from 0 to 5. Choose `binompdf`

from
the `DISTR`

menu, giving (5, .25) for the (n, p) arguments. Store this
in *L*_{2}. Now choose `binomcdf`

and supply the same arguments.
Store this in *L*_{3}. Now go to `STAT Edit`

to look at the lists.

The pdf column, *L*_{2}, contains the probabilities *P*(*X*=*x*), while the cdf
column, *L*_{3}, contains
.

(a)(1) *P*(*X*=5) can be found in *L*_{2}, the last entry:
`9.8E-4 = .00098.`

(a)(2)
can be found either by adding
*P*(*X*=4) + *P*(*X*=5) from
*L*_{2}, which is
.01465 + .00098 = .01563, or by taking
from *L*_{3}.

(a)(3)
*P*(*X*=0) = .2373, from *L*_{2}.

(a)(4)
,
from *L*_{3}.

Part (d) can be done easily with the calculator.

(d) Now we have n=50, p=.25, and we want
.
Here, let's just
get the cdf list, since adding up the pdf column from 30 to 50 would be almost
as tedious as doing the problem by hand!

Put the sample space {0, 1, ..., 50} in *L*_{4}. Put the cdf list in
*L*_{5}. Again, go to `STAT Edit`

to look at the lists. Scroll down
so that the "30" entry is in view.

.
So the probability is zero!
Not much chance of passing the exam with this strategy. Actually, this
probability isn't *exactly* zero, but it is very close to zero. The
TI-83 ran out of significant digits to display. This will be illustrated
when we use Excel/PHStat to do this problem next.

**Lastly, let's see how parts (a) & (d) can be done in Excel/PHStat.**

Visit www.stat.wmich.edu/s216/phstat.html to learn how to use the
Excel/PHStat software.

Choose *Probability Distributions* from the PHStat menu, then pick
*Binomial*.

For part (a), enter 5 for the *Sample Size*. The *Probability
of Success* should be .25, then enter 0 and 5 for *Outcomes From:
and To:* , respectively. Enter an *Output Title* if you wish, and
put a checkmark in the box next to *Cumulative Probabilities*.

On the *Binomial* sheet, notice that the mean and standard deviation are
given. These match the results we obtained by hand. Column B contains
the possible outcomes, C lists the pdf or *P*(*X*=*x*) probabilities (*L*_{2}
when we were using the TI-83), D lists the cdf or
probabilities (*L*_{3} from before), and columns E, F, and G contain the
probabilities from other ways to list them. Slightly overkill, but we can
make use of column G in this problem.

(a)(1)
*P*(*X*=5) = .000977, from column C.

(a)(2)
,
from column G.

(a)(3)
*P*(*X*=0) = .237305, from column C.

(a)(4)
,
from column D.

(d) Re-do the `Binomial Probability Distribution`

dialog box with
n=50, p=.25, `Outcomes`

from 0 to 50, and be sure the
`Cumulative Probabilities`

box is checked.

Scrolling down in order to view the probabilities when x=30, we see:

Now, can be found in column G.