15
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##
