Question

$100$ boys are randomly divided into two groups containing $50$ boys each. The probability that the two tallest boys are in different groups is

• A. $\frac{50}{99}$
• B. $\frac{49}{99}$
• C. $\frac{25}{99}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{50}{99}$

Let's say there are $2$ teams A & B

Boy 1 winds up on team A, say. That leaves $49$ open slots on Team A and $50$ on team B.

If Boy $2$ has equal probability of occupying any given one of these open slot then

the chance that the other of the tallest boys will be on team B = $\frac{50}{(49+50)}=\frac{50}{99}$

I think$\frac{50}{99}$ should be the probability that the two tallest boys are in different teams.