Question

Multiple-choice questions each have five possible answers (a, b, c, d, e), 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) (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) P(CWW) (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.)

Answer 1

Solution

given : \(P(\) correct \()\) or \(P(C)=\frac{1}{5}, P(\) wrong \()\) or \(P(W)=1-\frac{1}{5}=\frac{4}{5}\)

a. \(P(W W C)=\frac{4}{5} * \frac{4}{5} * \frac{1}{5}=\frac{16}{125}\)

b. \(P(W W C)=\frac{4}{5} * \frac{4}{5} * \frac{1}{5}=\frac{16}{125}\)

\(P(W C W)=\frac{4}{5} * \frac{1}{5} * \frac{4}{5}=\frac{16}{125}\)

\(P(C W W)=\frac{1}{5} * \frac{4}{5} * \frac{4}{5}=\frac{16}{125}\)

c. \(P(\) exactly one correct answer \() \Rightarrow P(W W C)+P(W C W)+P(C W W)\)

\(P(\) exactly one correct answer\() \Rightarrow \frac{16}{125}+\frac{16}{125}+\frac{16}{125}=\frac{48}{125}\)