Changed first part with a longer explanation

src/chapters/1/sections/naive_definition_of_probability/problems/26.tex

@@ -1,9 +1,15 @@
 \begin{enumerate}[label=(\alph*)]
-\item  When sampling with replacement, the probability of any sample of size 
-$1000$ is $$\frac{1}{K^{1000}}$$ where $K$ is the size of the population. 
-However, if sampling is done without replacement, then the probability is 
-$$\frac{1}{K(K-1) \dots (K-1000+1)}$$ which is different from the result in 
-the birthday problem.
+\item  The problem is isomorphic (having same structure) to the birthday
+problem. When sampling with replacement, each person corresponds to 
+a date in the birthday problem, and the size of sample corresponds to
+the number of people in birthday problem. Hence, taking a random
+sample of $1000$ from a population of a million corresponds to asking
+a thousand people their birth date where there are a total of a million
+dates. Number of ways to take such a sample is $K^{1000}$ where $K$ is
+size of population.
+Similarly, number of ways to take sample without replacement corresponds
+to number of ways of having no birthday match in that situation:
+$K(K-1) \dots (K-1000+1)$
 
 \item $$P(A) = 1 - P(A^{c}) = 1 - \frac{K(K-1) \dots (K-1000+1)}{K^{1000}}$$ 
 where $K = 1000000$.