We have Black cards and Red cards. We have 6 Black cards. We add Red cards until the probability of drawing two Red cards without replacement is #1/2#. How many Red cards do we need?

1 Answer
MathFact-orials.blogspot.com
Jan 12, 2017

15

Explanation:

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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