How many red cards must be added to 6 black cards so that the probability of drawing 2 red cards, without replacement, is 1/2?

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15

We feel (B)noncommunication cards and (R)ed cards. We initiate delay ##B=6## and absence to add R so that the odds of project 2 R, delayout renovation, is ##1/2##. How sundry R must we add?

The odds of project an R on a separate attract is:

##R/(R+B)##

and so for copy if we feel ##R=B=6##, we'd feel the odds of project an R to be:

##6/(6+6)=6/12=1/2##

So now let's add that avoid attract into the mix. We've already attractn an R and so we feel one R close, so the aspect for the avoid attract is:

##(R-1)/((R-1)+B)##

Which resources that the two attracts smitten contemporaneously are:

##(R/(R+B))((R-1)/((R-1)+B))##

We recognize ##B=6## and the overall odds we absence is ##1/2##, so we feel:

##(R/(R+6))((R-1)/((R-1)+6))=1/2##

And we can now explain for R:

##(R(R-1))/((R+6)(R+5))=1/2##

##(R^2-R)/(R^2+11R+30)=1/2##

we can now perverse multiply:

##2(R^2-R)=R^2+11R+30##

##2R^2-2R=R^2+11R+30##

##R^2-13R-30=0##

##(R-15)(R+2)=0##

##R=15,-2##

Since we can't add disclaiming R, we feel ##R=15##

Let's curb this defense.

The odds of the two attracts is:

##15/21 xx 14/20=210/420=1/2##

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