Probability
Aptitude

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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

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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 


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