In any case,\[40\%\] students study mathematics; \[25\%\] study biology and \[15\%\] study both mathematics and biology. One student is selected at random. Find the probability that he studies mathematics if it is known that he studies biology.

Hint: In order to find the probability of a randomly picked student, first we will be finding out the probability of the students studying mathematics only and then the probability of students studying biology only. Then we must find out the probability of students studying both the subjects. Then finally, we must consider two cases i.e. students studying both biology and mathematics and also students studying only biology. Solving this will give us the required value.

Complete step-by-step solution:
Now let us briefly talk about probability and its types. Probability can be defined as a chance of a particular event to occur from a set of events. The range of probability is between \[0\] and \[1\]. There are three types of probability. They are: theoretical probability, experimental probability and axiomatic probability. We can define an event as something that takes place.
Now let us start solving our problem.
Let us consider two cases as P,Q where P represents the students studying mathematics and Q would be the case of students studying biology.
Hence, we can say that \[P\cap Q\] will be the case of students studying both mathematics and biology.
So we are given that,
Probability of students studying mathematics\[=P(P)=40\%=\dfrac{40}{100}=\dfrac{2}{5}\]
Probability of students studying biology \[=P(Q)=25\%=\dfrac{25}{100}=\dfrac{1}{4}\]
Probability of students studying both mathematics and biology \[=P\left( P\cap Q \right)=15\%=\dfrac{15}{100}=\dfrac{3}{20}\]
Now,
The probability that the student studies mathematics if it is known that he studies biology will be
\[=P\left( P/Q \right)=\dfrac{P\left( P\cap Q \right)}{P\left( Q \right)}=\dfrac{3/20}{1/4}=\dfrac{3}{5}\]

Note: We are considering the intersection of students studying both mathematics and biology because the students are considered to be common. The common error could be not considering the intersection of probabilities in this case. The intersection case should not be considered as a separate probability.