(sum_{n=1}^{199}(n)) is written on cards. What is the probability of drawing a card with an even number written on it?


there is some typing mistake in the question,can you please edit it again


∑1n=1(1)=1∑n=11(1)=1—> odd

∑2n=2(1)=3∑n=22(1)=3—> odd

∑3n=3(1)=6∑n=33(1)=6—> even

∑4n=4(1)=10∑n=44(1)=10—> even

∑5n=1(5)=15∑n=15(5)=15—> odd

∑6n=1(6)=21∑n=16(6)=21—> odd

∑7n=1(7)=28∑n=17(7)=28—> even…. and so on

We see that we have 2 odds with evens.

This will go on till the cards having ∑196∑196 gives us 98 even and 98 odds numbers.

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

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