Binomial Probability Formula. In Exercises 13 and 14, answer the questions designed to help understand the rationale for the binomial probability formula.

Guessing Answers Standard tests, such as the SAT, ACT, or Medical College Admission Test (MCAT), typically use multiple choice questions, each with five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to the first three questions.

a. Use the multiplication rule to find the probability that the first two guesses are wrong and the third is correct. That is, find P(WWC), where W denotes a wrong answerand C denotes a correct answer.

b.Beginning with WWC, make a complete list of the different possible arrangements of two wrong answers and one correct answer, then find the probability for each entry in the list.

c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made?

A multiple choice-based exam is considered with five options, out of which one is correct.

Three such questions are answered.

a.

Here, W denotes a wrong answer, and C denotes a correct answer.

The probability of guessing a correct answer is computed below:

$$\begin{gathered} P\left(C\right)=\frac{1}{5} \\ =0.2 \end{gathered}$$

The probability of incorrectly guessing the answer is computed below:

$$\begin{gathered} P\left(W\right)=1-P\left(C\right) \\ =1-0.2 \\ =0.8 \end{gathered}$$

The probability of guessing the first two wrong answers and the third correct answer when a total of three questions are answered is computed below:

$$\begin{gathered} P\left(WWC\right)=P\left(W\right)P\left(W\right)P\left(C\right) \\ =\left(0.8\right)\left(0.8\right)\left(0.2\right) \\ =0.128 \end{gathered}$$

Thus, $P\left(WWC\right)=0.128$.

b.

Consider the following arrangements of answers to three questions when exactly two of them are wrong and one is correct:

$$\begin{gathered} WWC \\ WCW \\ CWW \end{gathered}$$

P(WWC) is equal to 0.128.

Now,

$$\begin{gathered} P\left(WCW\right)=P\left(W\right)P\left(C\right)P\left(W\right) \\ =\left(0.8\right)\left(0.2\right)\left(0.8\right) \\ =0.128 \end{gathered}$$

Thus, P(WCW) is equal to 0.128.

Now,

$$\begin{gathered} P\left(CWW\right)=P\left(C\right)P\left(W\right)P\left(W\right) \\ =\left(0.2\right)\left(0.8\right)\left(0.8\right) \\ =0.128 \end{gathered}$$

Thus, P(CWW) is equal to 0.128.

c.

The probability of exactly onecorrect answer is computed as follows:

$$\begin{gathered} P\left(1correctanswer\right)=P\left(WWC\right)+P\left(WCW\right)+P\left(CWW\right) \\ =0.128+0.128+0.128 \\ =0.384 \end{gathered}$$

Therefore, the probability of getting exactly one correct answer is equal to 0.384.