Question

multiple - choice questions each have 3 possible answers, one of which is correct. assume that you guess the answers to 4 such questions. use the multiplication rule to find the probability that the first three guesses are wrong and the fourth is correct. that is, find (p(wwwc)), where c denotes a correct answer and w denotes a wrong answer. (round answer to 4 decimal places) (p(wwwc)=0.0988) what is the probability of getting exactly one correct answer when 4 guesses are made? (round answer to 4 decimal places) (p(\text{exactly one correct answer}) = 0.2500)

Explanation:

Step1: Calculate probability of wrong and correct answer

The probability of a wrong answer $P(W)=\frac{2}{3}$ since there are 2 wrong answers out of 3. The probability of a correct answer $P(C)=\frac{1}{3}$ as there is 1 correct answer out of 3.

Step2: Use multiplication rule for independent events

The events of answering each question are independent. For $P(WWWC)$, we multiply the probabilities of each event: $P(WWWC)=P(W)\times P(W)\times P(W)\times P(C)=\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}=\frac{8}{81}\approx0.0988$.

Step3: Calculate probability of exactly one correct answer

We use the binomial probability formula $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n = 4$ (number of trials/questions), $k = 1$ (number of correct answers), $p=\frac{1}{3}$ (probability of success/correct answer). The binomial coefficient $C(n,k)=\frac{n!}{k!(n - k)!}$, so $C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4$. Then $P(X = 1)=4\times(\frac{1}{3})^{1}\times(1-\frac{1}{3})^{4 - 1}=4\times\frac{1}{3}\times(\frac{2}{3})^{3}=4\times\frac{1}{3}\times\frac{8}{27}=\frac{32}{81}\approx0.3951$.

Answer:

$P(WWWC)\approx0.0988$
$P(\text{exactly one correct answer})\approx0.3951$