91Ó°ÊÓ

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.