A chocolate gift box contains 15 chocolates six are Five Star, four are Fruit n Nut, five are Dairy Milk. After I have eaten the first chocolate, a Fruit n Nut, I pick another one. The probability that I pick a Fruit n Nut again is
\[\begin{align}
  & a.\dfrac{4}{15} \\
 & b.\dfrac{1}{3} \\
 & c.\dfrac{1}{5} \\
 & d.\dfrac{3}{14} \\
\end{align}\]

Hint: We will use the concept of probability to solve this question. We will use the formula, $ \text{Probability=}\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} $ . In this question, the selection is done without replacement, so the total number of outcomes and the favourable will decrease after each turn. We will get the total number of outcomes as (15 - 1) and the number of favourable outcomes as (4 - 1).

Complete step-by-step answer:
It is given in the question that a chocolate gift box contains 15 chocolates six are Five Star, four are Fruit n Nut, five are Dairy Milk. And after I have eaten the first chocolate, a Fruit n Nut, I pick another one. And we have to find the probability that I pick a Fruit n Nut again.
So out of the 15 chocolates I already ate one chocolate and we have to find the probability that I pick another Fruit n Nut from the remaining 14 chocolates.
Hence, the total number of outcomes will be = 14.
And, we have been given that the number of Five Star chocolates in the box = 5.
And, the number of Fruit n Nut = (4 - 1) = 3. Since, already 1 Fruit n Nut is eaten.
And the number of Dairy Milk chocolates = 5.
Now, we know that the formula of probability is given by, $ \text{Probability=}\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} $
So, in our case, the favourable outcome is that the second chocolate I pick is also a Fruit n Nut. And we know the number of these favourable outcomes is 3. And we have also found that the total number of outcomes is 14. So, on substituting these value in the formula of probability, we get,
 $ \text{Probability=}\dfrac{3}{14} $
Therefore, we get that the probability that I pick a Fruit n Nut again is $ \dfrac{3}{14} $ .
Hence, option (d) is the correct answer.

Note: The most possible and the common mistake that the students can make in this question is that, they may take the number of favourable outcomes as 4 instead of 3 and also they may take the total number of outcomes as 15 instead of 14, and hence, will get the answer as $ \dfrac{4}{15} $ , which is actually option (a), and this is the wrong answer. So, the students must note that as already 1 chocolate is eaten, the total outcomes will become 14 and the number of Fruit n Nut will become 3 as the one eaten was Fruit n Nut.