Probability that the number on cards with triangular numbers is even. Q: Summation(n) ($=1+2+\dots+n = n(n+1)/2$) is written for $n=1$ to $n=199$ on cards. What is the probability of drawing a card with an even number written on it?
Ans:it's going patternwise like first two is odd and next 2 is even
so I am getting it $99/199$.But the stated answer is $97/199$.
 A: Well as Aryabhata said, what is "summation(n)" ?
If it is the summation of all integers from $1$ to $n$, then what you are looking for is what is the condition on $n$ for it to be even.
Well, we know that your summation is equal to $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$
For it to be even means it has 2 as a divisor, so $\exists p\in\Bbb{N}\space s.t.\space \frac{n(n+1)}{2}=2p$ which in turn gives: $\exists p\in\Bbb{N}\space s.t.\space n(n+1)=4p$.
And this can mean two things:


*

*Case 1: $2$ is a divisor of both $n$ and $(n+1)$.

*Case 2: $4$ is a divisor of either $n$ or $(n+1)$.


However one of $n$ and $(n+1)$ is odd and the other even so case 1 is impossible. Which leaves us with case 2.
You can now say that if $S$ is the set of all $n$ that verify "summation(n) is even" then:
$S=\{n\in [1,199]:4|n\space or\space4|(n+1)\}$
Now think about the bounds of your interval. If you were going from 1 to a multiple of 4 (let's say 200 for example) then you would have half of them (100) odd and half even. But now you're removing one integer (200) which has an even summation, so you remove one even from 100 evens so you get 99 evens out of 199.
The stated answer is just wrong ;)
