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# Probability of a deck of Red and Blue cards

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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?

Mar 28, 2015

#3
+33151
+10

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

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Mar 28, 2015

#1
+33151
+10

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.

Mar 28, 2015
#2
+117845
+10

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

Mar 28, 2015
#3
+33151
+10

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

.

Alan Mar 28, 2015
#4
+124595
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Veey nice, Alan and Melody......!!!

Mar 28, 2015
#5
+117845
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Thanks Chris

Mar 29, 2015