<s>[INST] Probability of getting correct answer in a multi-choice question. 
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.
 [/INST] 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 
$$\frac{5}{15} = \frac{1}{3}$$ 
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