Charlie the Chimp is taking a multiple choice test. There are four questions on the test and there are three answer choices for each question. Only one answer choice is correct for each question. Since Charlie is a chimp, he will be guessing on each question. As a common fraction, what is the probability that he guesses exactly one right answer?
So for this, I know that since there are three possible answer and four questions, the denominater would be 3^4 or 81. Now for the numerator, I'm not really sure if this method is correct.
For unknown reasons (I just thought it seemed right) I did 4!/3....and then multiplied it by 4 because there are four problems, and got 32. So 32/81 is the answer...? Can someone provide a solution to find the numerater?
Oh wait, er I think I was overthinking a bit too much. Basically I think for the four questions, there are four cases.
To get the first question correct, and the rest of the questions need to be wrong, so the probability would be 1/3*2/3*2/3*2/3 or 8/81. I think I would multiply this by 4 because the question you get correct can be any question. So the answer is 32/81.....yay?
+118608 Charlie the Chimp is taking a multiple choice test.
There are four questions on the test and there are three answer choices for each question.
Only one answer choice is correct for each question.
Since Charlie is a chimp, he will be guessing on each question. As a common fraction, what is the probability that he guesses exactly one right answer?
4 independent trial. Probability of a correct answer each time is 1/3 so
P(exactly one correct)
\(= \binom{4}{1} * (\frac{1}{3})^1 * (\frac{2}{3})^3\\ =4 * \frac{1}{3}*\frac{8}{27}\\ =\frac{32}{81}\\ \)
