Socks in a drawer problem

[mathgeek7365]

My daughter is stumped, and so am I.

A drawer contains socks that are either blue or red. There are twice as many blue socks as red socks. If you select two socks from the drawer without replacement, the chance that you choose first a blue sock and then a red sock is 5/21. How many of each color sock was in the drawer before you removed any?

She knows that the probability to first draw a blue socks, then a red sock is 5/21.
The probability to draw a blue sock with the first draw is the number of blue socks
divided by the total number of socks. So, P(B)=b/n
She also knows that the probability to draw a red sock after one blue sock has been
taken out is the number of red socks divided by the total number of socks left after
you take one sock out. So, P(R)=r/(n-1). The probability of drawing a blue and then a red sock is shown as P(B,R)=(b/n)*(r/(n-1))=(5/21).
She also knows that there are twice as many blue socks as red socks, so b=2r.

She can replace b with 2r in the equation. P(B,R)=(2r/n)*(r/(n-1))=(5/21)

Is she correct so far? She would also like to know how to finish the problem. Thanks.

 

[Ackbach]

I would say you're on the right track. You do know that $2r=b$. You also, incidentally, know what $n$ is in terms of $r$ and $b$, right?

Your probability is given by
$$\frac{b}{n} \cdot \frac{r}{n-1}=\frac{5}{21}.$$
Now if you substitute everything in, and get it all in terms of one variable, you might be able to solve for one of the variables.