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)= what is the probability of getting exactly one correct answer when 4 guesses are made? (round answer to 4 decimal places) p(exactly one correct answer)=

Explanation:

Step1: Calculate probability of a wrong answer

Since there are 3 possible answers and 1 is correct, the probability of a wrong answer $P(W)=\frac{2}{3}$. The probability of a correct answer $P(C)=\frac{1}{3}$.

Step2: Calculate $P(WWWC)$

By the multiplication rule for independent events, $P(WWWC)=P(W)\times P(W)\times P(W)\times P(C)$. Substituting the values, we get $P(WWWC)=\frac{2}{3}\times\frac{2}{3}\times\frac{2}{3}\times\frac{1}{3}=\frac{8}{81}\approx0.0988$.

Step3: Calculate the probability of exactly one correct answer in 4 guesses

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 a 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 $P(X = 1)=4\times(\frac{1}{3})^{1}\times(\frac{2}{3})^{3}=4\times\frac{1}{3}\times\frac{8}{27}=\frac{32}{81}\approx0.3951$.

Answer:

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