the given question is this:
In the ii) part what should be the probability that the age is a prime number given that the age is greater than 15 years.
I have two contradicting methods of solving this problem , which would be the correct one.
Method 1: Using $n(f)/n(s)$ , it is known that the ages $16,17,19$ are greater than $15$ and $17 , 19$ are the prime numbers in that group so $F={17,19}$ and $S={16 ,17 , 19 }$ which yields us $2/3$
Method 2: Using the probability distribution it is known that $16$ has a probability of appearing to be $3/10$ out of the bunch and 17 has a probability of $2/10$ similarly $19$ has a probability of $2/10$ , using conditional probability we know P(PRIME given its greater than 15) $=\frac{\frac{2+2}{10}}{\frac{2+2+3}{10}}$ which yields $4/7$
Which method do you think would be the correct answer for question ii)
You have to sample among the students, not among their ages.
There are $3$ students who are $16$ years of age; $2$ students who are $17$; and $2$ who are $19$. Therefore, in all, there are $7$ students who could have been selected, given that the student actually selected was more than $15$ years old.
Among these $7$ students, only those who are $17$ or $19$ years old have ages that are prime numbers; thus, the desired probability is $4/7$.
The reasoning used in the first method of solution is incorrect because, as I stated at the beginning, you must sample among the students. The problem literally states:
It does not say "an age was selected at random." Not all of the possible ages are equally represented. To illustrate, suppose you have a special six-sided die, in which five of the six sides are labeled $1$ and the remaining sixth side is labeled $6$. Each side is equally likely to be rolled. What is the probability you roll a $1$? It is not $1/2$ simply because the only outcomes are $1$ and $6$.