Question

Let $S_n=\sum _{k=1}^n$ k denote the sum of the first n positive integers The numbers $S_1,S_2,S_3.....S_{99}$ are written on 99 cards.The probability of drawing a cards 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)

$\frac{49}{99}$

$S_n={\sum }_{k=1}^{n}k=\frac{k(k+1)}{2}$

$S_1=1,S_2=3,S_3=6,S_4=10,S_5=15,S_6=21$

Hence, the terms $3,4,7,8,11,12$

Hence the even terms are the terms which are of alternate consecutive number.

Let $k_1+k_2$ be the total even terms

$3+(k_1-1)4=99(k_1-1)4=96k_1-1=24k_1=25$

$4k_2=96k_2=24$
Total even terms $k_1+k_2=49$

Probability $=\frac{49}{99}$