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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.

AUse the multiplication rule to find ​P(WWC​), where C denotes a correct answer and W denotes a wrong answer.

​P(WWC​)=

BBeginning 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​)=

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

Verified Answer :

kudzordzifrancis
ASince 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.

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

The answer to each of the given options on probability are;

A) P(WWC) = ⁹/₆₄

B) P(WWC) = ⁹/₆₄

P(WCW) = ⁹/₆₄

​P(CWW​) =⁹/₆₄

C) P(one correct answer) = ²⁷/₆₄

Probability Results

We are told that there are four possible answers and only one is correct.

Thus,

Probability of correct answer is; P(C) = ¹/₄

Probability of wrong answer is; P(W) = ³/₄

A) P(WWC​) = P(W) × P(W) × P(C)

P(WWC) = ³/₄ × ³/₄ × ¹/₄

P(WWC) = ⁹/₆₄

B) The 3 different possible arrangements are;

WWC, WCW and CWW. Thus;

P(WCW) =  ³/₄ × ¹/₄ × ³/₄

P(WCW) = ⁹/₆₄

P(CWW) = ¹/₄ × ³/₄ × ³/₄

P(CWW) = ⁹/₆₄

C) To find out the probability of getting exactly one correct answer when three guesses are​ made is;

P(one correct answer) = P(WWC) +  P(WCW) + P(CWW)

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

P(one correct answer) = ²⁷/₆₄

Read more about probability results at; https://brainly.com/question/25870256

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