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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We have (B)lack cards and (R)ed cards. We start with ##B=6## and want to add R so that the odds of drawing 2 R, without replacement, is ##1/2##. How many R must we add?

The odds of drawing an R on a single draw is:

##R/(R+B)##

and so for example if we have ##R=B=6##, we’d have the odds of drawing an R to be:

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

So now let’s add that second draw into the mix. We’ve already drawn an R and so we have one R less, so the ratio for the second draw is:

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

Which means that the two draws taken together are:

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

We know ##B=6## and the overall odds we want is ##1/2##, so we have:

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

And we can now solve for R:

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

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

we can now cross 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 negative R, we have ##R=15##

Let’s check this answer.

The odds of the two draws is:

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

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