+0  
 
0
928
5
avatar+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?

 Mar 28, 2015

Best Answer 

 #3
avatar+30591 
+10

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

.

 Mar 28, 2015
 #1
avatar+30591 
+10

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

 Cards 1

 

 Cards 2

 

Now with replacement:

 Cards 3

 

 Cards 4.

 

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

 Mar 28, 2015
 #2
avatar+110080 
+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
avatar+30591 
+10
Best Answer

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

.

Alan Mar 28, 2015
 #4
avatar+111432 
0

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

 

  

 Mar 28, 2015
 #5
avatar+110080 
0

Thanks Chris

 Mar 29, 2015

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