Question

Consider a short multiple choice quiz with three items, and each item has four choices (only one of the choices is correct). Suppose that you are taking this quiz but you are completely and utterly unprepared for it. That means that you only option for the quiz is to guess the answers. Suppose you are thinking about the first item: what's the probability that you'll answer it incorrectly (hint: think about how many options there are, and how many of them are wrong)?

Now, list the "sample space" of all possible outcomes on this exam. Hint: here is one possible outcome: Correct, Correct, Wrong (or, CCW for short).

Using the format in #2 above, we actually haven't listed the entire sample space, because there are multiple ways to answer each item incorrectly! In order to calculate the probability of a particular outcome, we either need to list every single way to answer a problem wrong, or we can use the multiplication rule. These outcomes are all independent because guessing on problem #1 has no effect on your guess for problems #2 or #3. For example, P(Correct on the 1st, and Correct on the 2nd, and Wrong on the 3rd) = P(Correct on the 1st) * P(Correct on the 2nd) * P(Wrong on the Third). Find P(CCW).

Now, use your list from #2 and write the probability for each outcome next to it, using the same type of calcuation you did in #3.

Next, use your work from #4 to fill in the following table with the relevant probabilities, where x represents the number of items you answer correctly. Notice that the number of correct items will sometimes include more than one of the listed outcomes (what do you do with those numbers?)

x	P(x)
0
1
2
3

In #5 above, you should have a probability distribution (which means that all probabilities are between 0 and 1, and the sum of the probabilities is 1). Check to make sure this is correct, fix any errors, and then explain why this situation fits the criteria for a binomial distribution.

Using the table you made in #5, determine the probability of getting at most 2 items correct on the quiz.

We don't need to calculate each probability by hand. It can be done in a spreadsheet: Using excel or sheets, you can type the command "BINOM.DIST" (in sheets, there is no dot) to help compute these probabilities. For example, the command "=BINOM.DIST(0,3,.25,0)" tells you the probability of getting 0 correct (# of "successes") out of three (# of trials) where the probability of getting one item correct is .25 (probability of a "success"), and we want exactly that number of correct items ("0" at the end of the command refers to the fact that we aren't adding up any probabilities, that the "cumulative" feature of the command is turned off). Verify, using this command, that each of the values in your table above is correct.

Now, you can also check the calculation you did in #7, by typing the following command: "BINOM.DIST(2,3,.25,1)", which means you want to get two correct (# of successes) out of three items (# of trials), where the probability of getting one item correct is .25 (probability of a success), and we are adding up the probabilities for all values below 2 as well (the cumulative spot is turned on when you type "1").

Finally, create a histogram of the distribution and copy it into your response.

Answer 1

There are 3 items with 4 choices out of which only one is correct.

The probability of getting the correct answer is 1/4 = 0.25

and Probability of getting incorrect answer = 3/4 = 0.75

For sample space there are 2³ = 8 possible outcomes, 2 is there are 2 outcomes either correct or wrong and 3 is nothing but a total number of items.

The sample space is

s = {CCW, CWC, CWW, CCC, WWW, WWC, WCW, WCC)

Where C - correct and W - wrong.

X: number of items answered correctly.

The possible values of x are 0, 1, 2 and 3

Let us find the probability distribution for x

X = 0 means 0 are correct that means all are wrong

X = 1 means only 1 is correct out of 3, so there are three possibilities

Either first correct, second and third wrong OR first wrong, second correct and third wrong OR first two are wrong and the third is correct.

X = 2 means 2 are correct out of 3, there are also three possibilities

Either first 2 are correct and the third is wrong OR first and third are correct and second is wrong OR first is wrong and the last two are correct.

And X = 3 means all three are correct

Therefore, the probability distribution of x is:

X	P(X)
0	0.421875
1	0.421875
2	0.140625
3	0.015625

The all 4 probabilities in the above table are between 0 to 1 and the sum of probabilities is 1,

0.421875 +0.421875+0.140625+0.015625 = 1

Therefore, this is the probability distribution.

Here there are only two outcomes for each item either it is correct or wrong, the probability of success is 1/4 = 0,25 which is the same throughout the trails, also the number of trials = 3 which are independent on each other, therefore this random variable X follows binomial distribution with n = 3 and p = 0.25.

By using BINOM.DIST function is excel the probabilities are:

P(X = 0) =BINOM.DIST(0, 3, 0.25, 0) = 0.421875

P(X = 1) = BINOM.DIST(1, 3, 0.25, 0) = 0.421875

P(X = 2) = BINOM.DIST(2, 3, 0.25, 0) = 0.140625

P(X = 3) = BINOM.DIST(3, 3, 0.25, 0) = 0.015625

The histogram for the number of successes with their corresponding probabilities is,