What is the probability of getting 5 multiple-choice questions answered correctly, if for each question the probability of answering it correctly is 1/3.

The answer is 45/118, but I am unsure of how. Update: The book may have had the question worded incorrectly, because the answer stated is incorrect.

What i know: Each question has 3 selections, and the probability of getting one wrong is 2/3. I thought it would be as simple as the multiplicative rule, but there's more to it apparently. I think it has to do with a formula as you go along in the questions, but it states that each probability is equal, as a 1/3 chance.

Any suggestions?

  • $\begingroup$ You did not say how many questions are on the test! $\endgroup$ – André Nicolas Jan 31 '13 at 1:46
  • $\begingroup$ haha it's 9pm here and i'm going through a book of hundreds of questions! $\endgroup$ – Ben Sewards Jan 31 '13 at 1:55
  • $\begingroup$ @AndréNicolas 5 multiple-choice questions, right? (first sentence?) $\endgroup$ – apnorton Jan 31 '13 at 1:56
  • $\begingroup$ @anorton I think 5 is the number of questions answered correctly. $\endgroup$ – Patrick Li Jan 31 '13 at 1:58
  • $\begingroup$ @PatrickLi Oh. thx $\endgroup$ – apnorton Jan 31 '13 at 2:07

If there are $n$ questions on the test, then the probability of answering exactly $k$ correctly, if answers are chosen at random, is $$\binom{n}{k}\left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{n-k}.$$

And it is the Muliplication Rule. To make typing simpler, I will take $k=2$, $n=5$. Write $C$ for correct, $N$ for not correct.

By the Multiplication Rule, the probability of $CCNNN$ (first two right, next three wrong) is $(1/3)^2(2/3)^3$. But we can also get two right, three wrong in several other ways, like $CNCNN$, $CNNCN$, and so on. Each has probability $(1/3)^2(2/3)^3$.

How many such strings are there? We have to choose $2$ places from the $5$ available to put a $C$ into. There are $\binom{5}{2}$ ways to do this, giving total probability $\binom{5}{2}(1/3)^2(2/3)^3$.

The same reasoning gives the general formula quoted above.

Remark: If there are only $5$ questions on the test, and we want the probability of answering all $5$ correctly, the probability is $(1/3)^5$, which is nowhere near the number quoted.

Because the formula for the probability has only $3$'s in the denominator, any probability we compute will have to be of shape $\frac{m}{3^e}$, where $m$ and $e$ are non-negative integers. The number $\frac{45}{118}$ is not of that shape, so cannot be the answer for the problem, or for any closely related problem.

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  • $\begingroup$ It is impossible to get $45/118$ as OP mentioned. $\endgroup$ – Patrick Li Jan 31 '13 at 1:57
  • $\begingroup$ i have no idea why that was the answer in the book, investigation time -.- $\endgroup$ – Ben Sewards Jan 31 '13 at 1:59
  • $\begingroup$ @BenSewards You still didn't say the total number of questions. $\endgroup$ – Patrick Li Jan 31 '13 at 1:59
  • $\begingroup$ no total number of questions were of concern in the question. Maybe a total of 5? I got the same answer as above, so I guess I'll stick with that in the future when it is the case. $\endgroup$ – Ben Sewards Jan 31 '13 at 2:01

A simple bit of logic can quickly prove that 45/118 is incorrect. The chance of guessing one answer on its own is 1/3. 45/118 is slightly more than 1/3 which means that guessing all 5 correctly is more probable than guessing only one correctly. This is obviously untrue as guessing each additional answer is increasingly less likely.

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