Question

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?

Answer 1

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`