Question

This Test: 25 pls poss Question Help Multiple-choice questions each have four possible answers (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 denotes a correct answer and W denotes a wrong answer, PWWC)-(Type an exact answer) 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) - see above P(WCW) - PICWW) = (Type exact answers) C. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made? (Type an exact answer Enter your answer in each of the answer boxes

Answer 1

Let us consider a multiple choice question with each possible four answer where one of the option (a), (b), (c), and (d) is correct.

Since there are four option, so total number of cases is 4 and only one option is correct so number of favourable case is one.

Total number of Cases = 4.

Favourable Cases=1.

We know that the probability is given by

$Probability=p=\frac{Favourable \ Cases}{Total \ number \ of \ Cases}.$

Let C denotes the correct answer and W denotes the wrong answer.

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

Since only one option is correct so remaining three would be wrongly answered.

  

P(W) = P(Wrong answer) 3 1

a) For obtaining the value of P(WWC), we make use of the concept that out of three question answered two of them is wrongly answered and last one is answered correctly.

All the questions are independently answered so the probability of each question answered does not depend on one another. Therefore we have

$P(WWC)=Probability\ that\ first\ two\ answer\ is\ wrongly \ answered \ and \ \\ third \ one \ is \ correctly \ answered. \ Thus \\ P(WWC)=P(W)*P(W)*P(C) \\ P(WWC)=\frac{3}{4}*\frac{3}{4}*\frac{1}{4}=\frac{9}{64}=0.1406.$

Therefore the probability is 0.0468.

b) Different possible arrangements of one correct answer out of three question answered  is CWW, WCW, WWC.

P(WWC)=Probability that third one is answered correctly and all the first two are wrongly answered which is given from the previous one is 0.0468.  

For P(WCW) the probability is

$P(WCW)=Probability\ that\ first\ and \ third \ question \ is\ wrongly \ answered \\ and \ second \ one \ is \ correctly \ answered. \ Thus \\ P(WCW)=P(W)*P(C)*P(W) \\ P(WCW)=\frac{3}{4}*\frac{1}{4}*\frac{3}{4}=\frac{9}{64}=0.1406.$

For P(CWW) the probability is given by

$P(CWW)=Probability\ that\ second \ and \ third \ question \ is\ wrongly \ answered \\ and \ first \ one \ is \ correctly \ answered. \ Thus \\ P(CWW)=P(C)*P(W)*P(W) \\ P(CWW)=\frac{1}{4}*\frac{3}{4}*\frac{3}{4}=\frac{9}{64}=0.1406.$

Therefore the value of P(WCW)=0.1406, P(CWW)=0.1406 and P(WWC)=0.1406.

c) The probability of getting exactly one correct answer when three guesses are made is obtained by considering all the probability of P(WWC), P(WCW), P(CWW).

Probability of one answer out of three guessed answer is correct = P(WWC)+P(WCW)+P(CWW).

Since, P(WWC)=0.1406, P(CWW) = 0.1406, P(WCW)=0.1406.

Then the probability of one answer correctly answered out of the three guessed answer =0.1406+0.1406+0.1406

=0.4218.