# Difficult Probability Solved QuestionAptitude Discussion

 Q. (sum_{n=1}^{199}(n)) is written on cards. What is the probability of drawing a card with an even number written on it?
 ✖ A. 1/2 ✖ B. 97/200 ✔ C. 99/199 ✖ D. 97/199

Solution:
Option(C) is correct

$\sum_{n=1}^{1}(1)=1$---> odd

$\sum_{n=2}^{2}(1)=3$---> odd

$\sum_{n=3}^{3}(1)=6$---> even

$\sum_{n=4}^{4}(1)=10$---> even

$\sum_{n=1}^{5}(5)=15$---> odd

$\sum_{n=1}^{6}(6)=21$---> odd

$\sum_{n=1}^{7}(7)=28$---> even…. and so on

We see that we have 2 odds with evens.

This will go on till the cards having $\sum196$ gives us 98 even and 98 odds numbers.

For $n=197$ and $n=198$ It is odd and for $n=199$ it is even.

Thus, required probability $= (98+1)/199=$ 99/199