Question

Let $S_n=\sum _{k=1}^{n}k$ denote the sum of the first $n$ positive integers. The numbers $S_1,S_2,S_3,\dots ,S_{99}$ are written on $99$ cards. The probability of drawing a card with an even number written on it is

• A. $\frac{1}{2}$
• B. $\frac{49}{100}$
• C. $\frac{49}{99}$
• D. $\frac{48}{99}$

Answer 1

Answer: (C)

$\frac{49}{99}$

Now here $S_1=1,S_2=3,S_3=6,S_4=10,S_5=15,S_6=\mathrm{21....}$ we see that every $2$ terms are odd,even alternately.

$S_{97,}{S}_{98}$ will be odd and $S_{99}$ will be even.

so there will be one odd term than even term,

so total terms $=99$

and even terms$=49$

Required probability $\frac{49}{99}$