Probability of a deck of Red and Blue cards

Looks like you and I agree for the "with replacement" case Melody.

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Eventually I might get this right!  Here's my latest attempt:

Now with replacement:

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So, with and without replacement cases require 16 red cards.

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  

Veey nice, Alan and Melody......!!!

Thanks Chris