# Probability of two events

Suppose that a friend gives us the chance to participate in a game. The game is as follows. There is a bucket containing $25$ green tokens and $10$ red tokens. We are to blindly reach into the bucket and randomly choose one token. Keeping this token (i.e. the chosen token is not returned to the bucket) we are to once again reach in blindly and randomly retrieve another token. If both of the tokens we have chosen are green then we win the game. Otherwise we lose. Define the two events below as follows:

• $G_1$ = first token selected is green • $G_2$ = second token selected is green.

Suppose that you would like to increase the probability that you win the game and will propose a new total number of green and red tokens to be used. You would like your probability of winning the game to be at least $0.6$ (i.e. $P(G1 \cap G2) \geq 0.6$). What is the smallest total number of tokens needed (number of green tokens plus the number of red tokens) to achieve this? Please note that there must be at least one red token to play this game. For this total number of tokens, how many are green and how many are red?

Let's say you have $g$ green tokens and $r$ red tokens. The the probability you win is $$p=\frac{g}{g+r}\frac{g-1}{g+r-1}$$ The probability is strictly decreasing in $r$ so we need to minimize it. So set $r=1$. Then, $$p = \frac{g-1}{g+1}$$ The smallest value of $g$ such that $p \geq 0.6$ is $g=4$.