Question

Multiple-choice questions each have four possible answers left parenthesis a , b , c , d right (a, b, c, d)​, one of which is correct. Assume that you guess the answers to three such questions.
a. Use the multiplication rule to find ​P(WWC​), where C denotes a correct answer and W denotes a wrong answer.
​P(WWC​)=

b. Beginning with WWC​, make a complete list of the different possible arrangements of one correct answer and two wrong answers​, then find the probability for each entry in the list.
​P(WWC​)minus−see above
​P(WCW​)=
nothing
​P(CWW​)=

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

Answer 1

P (WWC) = 3/4 * 3/4 * 1/4 = 9/64
P (WCW) = 9/64
P (CWW) = 9/64

P (exactly one correct answer) = 9/64+9/64+9/64 = 27/64

Answer 2

A) Since there are four multiple choice questions Which has one correct answer. The probability of choosing a correct answer is

$P(C) = \frac{1}{4}$

The probability of choosing wrong answer is

$P(W) = \frac{3}{4}$

Using the multiplication rule
$P(WWC) = P(W) \times P(W) \times P(C) \\ \\ P(WWC) = \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} \\ \\ P(WWC) = \frac{9}{64}$

b) If you guess answers to three of the questions, then the possibilities of getting one correct answer are:

Either the first two are wrong and the third one is correct. This will give the arrangement;

$WWC$

Or the first is wrong the second one is correct and the last one is wrong. This will give the arrangement,

$WCW$

Or the first one is correct and the last two are wrong. This will give the arrangement,

$CWW$
.

$P(WWC) = P(W) \times P(W) \times P(C) \\ \\ P(WWC) = \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} \\ \\ P(WWC) = \frac{9}{64}$

$P(WCW ) = P(W) \times P( C ) \times P(W) \\ \\ P(WCW) = \frac{3}{4} \times \frac{1}{4} \times \frac{3}{4} \\ \\ P(WCW) = \frac{9}{64}$

$P(CWW ) = P(W) \times P( C ) \times P(W) \\ \\ P(CWW) = \frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} \\ \\ P(CWW) = \frac{9}{64}$

c) The probability of getting one correct answer is either the first one is correct or second is correct or third is correct.

$P(One \: Correct)= P(CWW) \: or P(WCW) \: or \: P(WWC) \\ \\ P(One \: Correct)= P(CWW) \: + P(WCW) \: + \: P(WWC) \\ \\ P(One \: Correct)= \frac{9}{64} + \frac{9}{64} + \frac{9}{64} \\ \\ P(One \: Correct) = \frac{27}{64}$