In a multiple choice question there are 4 alternative answers of which 1, 2, 3, or all may be correct. A candidate decides to tick answers at random. If he is allowed upto 5 chances to answer the question, the probability that he will get the marks in the question is?

It is equally likely for 1, 2, 3, or all to be correct so probability is 1/4.

Case1 when only 1 option is correct.
Since the candidate is allowed 5 chances, probability of getting correct answer is 1.

Case2 when 2 options are correct.
Toltal ways in which 2 options can be correct is $\left(^4_2\right)$, which is 6. Out of these only 1 is correct. So probability of selecting correct answer is 1/6. Since he has 5 chances the probability of getting marks is $1 - \left(\frac{5}{6}\right)^5$

Case3 when 3 options are correct.
Total ways in which 3 options can be correct is $\left(^4_3\right)$, which is 4. Since he has 5 chances, probability of getting correct answer is 1.

case4 when all options are correct.
Only one way in which all can be correct. Probality of getting marks is 1.

So answer should be $\left(\frac{1}{4}\times1\right) + \frac{1}{4}\times\left(1 - \left(\frac{5}{6}\right)^5\right) + \left(\frac{1}{4}\times1\right) + \left(\frac{1}{4}\times1\right)$.
Which is indeed wrong.

A friend of mine did this question as follows.
Total options: $\left(^4_1\right) + \left(^4_2\right) + \left(^4_3\right) + \left(^4_4\right)$, i.e 15. Since he has 5 chances to answer, probability would be 5/15. I know this is wrong (or not?) but beacause my textbook says answer is 1/3 I couldn’t argue.

Please help. Thanks in advance.

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    $\begingroup$ You are assuming that the candidate knows how many answers are correct, e.g. in the case that only one is correct, the candidate will not mark multiple questions (which would, in my understanding, be considered a wrong answer in a multiple choice test...). Is that intended? Furthermore, the fact that every case has the same chance of 1/4 is given in the question? If not, why can you assume it? I agree with your friend, but he starts with other assumptions, therefore it is important to know why you make them. $\endgroup$ – Dirk Oct 19 '17 at 10:26
  • $\begingroup$ @DirkLiebhold I agree with your argument about my solution that I made wrong assumptions. Can you please explain why my friend's solution is right. $\endgroup$ – Voneone Oct 19 '17 at 10:39
  • $\begingroup$ Your friend is correct. $\endgroup$ – N. F. Taussig Oct 19 '17 at 11:12
  • $\begingroup$ @N.F.Taussig Can you please add answer explaining the same? $\endgroup$ – Voneone Oct 19 '17 at 11:23
  • $\begingroup$ I am unable to understand how having 5 chances gives use 5 favourable events. Is it because we are adding probability of all five cases with each case having probability 1/15, i.e 1/15 + 1/15 + 1/15 + 1/15 + 1/15 $\endgroup$ – Voneone Oct 19 '17 at 11:30

There are four choices, each of which may be correct or incorrect. However, not all of them may be incorrect. Therefore, there are $2^4 - 1 = 15$ possible ways to answer the question. The candidate is allowed five guesses. That means the candidate chooses $5$ of the $15$ possible answers. The candidate receives credit if one of those guesses is correct, so the probability that the candidate guesses the correct answer is $$\frac{5}{2^4 - 1} = \frac{5}{15} = \frac{1}{3}$$

Your friend counted the number of possible answers in a different way. We have the option of choosing one, two, three, or four of the answers. There are $\binom{4}{k}$ ways to choose exactly $k$ of the four answers. Hence, the number of possible ways to answer the question is $$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$$

Edit: To understand why the probability is the number of guesses divided by the number of answers, suppose that you are given $k$ guesses and $n$ answers, of which only one is correct. There are $\binom{n}{k}$ ways to make $k$ guesses. If you select the correct answer with one of those $k$ guesses, then you also select $k - 1$ of the $n - 1$ incorrect answers. Therefore, your probability of selecting the right answer among your $k$ guesses is \begin{align*} \frac{\dbinom{1}{1}\dbinom{n - 1}{k - 1}}{\dbinom{n}{k}} & = \frac{\dfrac{(n - 1)!}{(k - 1)![(n - 1) - (k - 1)]!}}{\dfrac{n!}{k!(n - k)!}}\\ & = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{k!(n - k)!}{n!}\\ & = \frac{(n - 1)!}{n!} \cdot \frac{k!}{(k - 1)!}\\ & = \frac{k}{n} \end{align*} In our problem, $k = 5$ and $n = 15$, so the probability of obtaining the correct answer is $$\frac{5}{15} = \frac{1}{3}$$

  • $\begingroup$ We have 5 guesses, but that does not mean 5 ways out of 15 ways are correct. So why is the answer 5/15? Am I missing something here? $\endgroup$ – Voneone Oct 19 '17 at 11:59
  • $\begingroup$ There is only one correct answer. However, the candidate has five attempts. If one of those five attempts is correct, the candidate gets credit for the answer. In those five attempts, the candidate can choose five of the fifteen possible answers. Therefore, a third of the time, the candidate will choose the correct answer in one of those five attempts. $\endgroup$ – N. F. Taussig Oct 19 '17 at 12:16
  • $\begingroup$ Your explanation , "the candidate can choose five of the fifteenth possible answer", created more confusion than helping. You don't choose options like this. Can anyone explain the steps in more detail or with some other example. $\endgroup$ – Voneone Oct 19 '17 at 18:09
  • $\begingroup$ There are $\binom{15}{5}$ ways to choose five of the fifteen answers. If you choose the correct answer, you must also choose four of the fourteen incorrect answers. Therefore, the probability that you choose the correct answer is $$\frac{\binom{1}{1}\binom{14}{4}}{\binom{15}{5}} = \frac{5}{15} = \frac{1}{3}$$ In general, if you have $k$ choices, then the probability that you obtain the correct answer is $$\frac{\binom{1}{1}\binom{14}{k - 1}}{\binom{15}{k}} = \frac{k}{15}$$ as you should check. Notice that the answer is the number of guesses divided by the number of possible answers. $\endgroup$ – N. F. Taussig Oct 19 '17 at 19:11
  • 1
    $\begingroup$ Thanks. It's clear now. $\endgroup$ – Voneone Oct 19 '17 at 19:37

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