Question

In a box, there are n tickets, each has a positive integer on it. We choose two tickets with replacement from the box. Show that P(sum of the two tickets is even) 2 1/2.

Answer 1

We draw two tickets with replacement from the box with n tickets.
Let's first look at what sum of two positive integers yield (in terms of even and odd result)
Even + Even = Even
Even + Odd = Odd
Odd + Odd = Even
Suppose n is even:
Then there are n/2 odd tickets and n/2 even tickets
P(Even ticket) = 0.50
P(Odd ticket) = 0.50
P(Sum of two tickets even) = P(First ticket even).P(Second ticket even) + P(First ticket odd).P(Second ticket odd) = 0.50 x 0.50 + 0.50 x 0.50 = 0.50
P(Sum of two tickets is Even) = 1/2
Suppose n is odd:
Then there are (n-1)/2 even tickets and (n+1)/2 odd tickets
P(Even ticket) = (n-1)/2n
P(Odd ticket) = (n+1)/2n
P(Sum of two tickets even) = P(First ticket even).P(Second ticket even) + P(First ticket odd).P(Second ticket odd):
= (n-1)² / 4n² + (n+1)² / 4n² = (n² + 1) / 2n² = (n²/2n²) + (1/2n²) = 1/2 + (1/2n²)
P(Sum of two tickets even) = 1/2 + (1/2n²)
i.e. P(Sum of two tickets even)  > 1/2

So, in case n is even: P(Sum of two tickets even) = 1/2 and in case n is odd P(Sum of two tickets even)  > 1/2,
Overall, P(Sum of two tickets even) ≥ 1/2