+5 I asked a similar question recently, but now I have a modification:
You have a deck of 60 cards: Red cards and Blue cards. You draw 6 cards. How many Red cards need to be in the deck so that there's a 50% chance there's AT LEAST 2 Red cards among the 6 you drew?
+33615 Eventually I might get this right! Here's my latest attempt:
Now with replacement:
.
So, with and without replacement cases require 16 red cards.
+118608 I am going to assume the cards are replaced each time.
Let x be the number of blue and 60-x be the number of red
P(at least 2 red) >=0.5
P(0 or 1 red) <=0.5
P(all blue)+P(1R and 5 blue) <=0.5
$$\\\frac{x^6}{60^6}+6C1\left[\frac{(60-x)}{60}\right]^1\left[\frac{x}{60}\right]^5\le0.5\\\\\\
\frac{x^6}{60^6}+6\left[\frac{(60-x)x^5}{60^6}\right]\le0.5\\\\\\$$
I used Wolfram|Alpha to solve this
http://www.wolframalpha.com/input/?i=%28x^6%2F60^6%29%2B6*%28%28%2860-x%29*x^5%29%2F60^6%29%3C0.5
$$\\x\le44\\\\
$So the must be less than 44 or equal to 44 blue cards$\\\\
$Which means that there must be 16 or more red cards.$ \\\\$$
check:
$$\frac{44^6}{60^6}+\frac{6*(60-44)*44^5}{60^6} = 0.4949$$ seems alright 
