# Thread: socks in a drawer probability problem

1. ## socks in a drawer probability problem

Hi all

I have a probability problem book and one problem looks at the solution of the following:

a drawer contains red socks and black socks. when two socks are drawn at random the probability that both are red is 0.5. how small can the number of socks in the drawer be? how small if the number of black socks is even?

the solution solves for the general case with the following with r red socks and b black socks:

1. r/(r+b) x r-1/(r+b-1)=0.5
then sets up the inequalities

2. (r/(r+b))^2 > 0.5 > (r-1/(r+b-1))^2

taking the square root and multiplying by (r+b) gives first inequality as

3. r > 1/(2^0.5) x (r+b)

and the next step takes it to:

4. r > 1/(2^0.5 - 1) x b

I do not follow why we are able to go from 3 to 4. can anyone help out please?

2. Hello, factfinder!

A drawer contains red socks and black socks.
When two socks are drawn at random, the probability that both are red is $\tfrac{1}{2}$
How small can the number of socks in the drawer be?
How small if the number of black socks is even?

The solution solves for the general case with $r$ red socks and $b$ black socks:

. . $1.\;\;\frac{r}{r+b}\cdot\frac{r-1}{r+b-1}\:=\:\frac{1}{2}$

Then sets up the inequalities:

. . $2.\;\;\left(\frac{r}{r+b}\right)^2 \:>\: \frac{1}{2} \:>\:\left(\frac{r-1}{r+b-1}\right)^2$

Taking the square root and multiplying by $(r+b)$ gives first inequality as:

. . $3.\;\; r\: >\:\frac{1}{\sqrt{2}}(r+b)$

and the next step takes it to:

. . $4.\;\; r \:>\: \frac{1}{\sqrt{2}-1}\,b$

I do not follow why we are able to go from 3 to 4.
They skippped some fancy footwork . . .

At (3) we had: . $r \:>\:\frac{1}{\sqrt{2}}(r+b)$

Multiply by $\sqrt{2}\!:\;\;\sqrt{2}\,r \:>\:r + b \quad\Rightarrow\quad \sqrt{2}\,r - r \:>\:b$

Factor: . $\left(\sqrt{2}-1\right)r \:>\:b \quad\Rightarrow\quad r \:>\:\frac{1}{\sqrt{2}-1}\,b$

3. Originally Posted by factfinder
Hi all

I have a probability problem book and one problem looks at the solution of the following:

a drawer contains red socks and black socks. when two socks are drawn at random the probability that both are red is 0.5. how small can the number of socks in the drawer be? how small if the number of black socks is even?

the solution solves for the general case with the following with r red socks and b black socks:

1. r/(r+b) x r-1/(r+b-1)=0.5
then sets up the inequalities

2. (r/(r+b))^2 > 0.5 > (r-1/(r+b-1))^2

taking the square root and multiplying by (r+b) gives first inequality as

3. r > 1/(2^0.5) x (r+b)

and the next step takes it to:

4. r > 1/(2^0.5 - 1) x b

I do not follow why we are able to go from 3 to 4. can anyone help out please?
(a) 4 is the smallest number of possible socks. 3 red and 1 black.

(b) If black socks are even numbered the smallest number is 21. 6 black and 15 red.

I don't know how to use the symbols program, so this is going to be messy. Set up this equation: Let x= r+b and use a hyper-geometric probability function with reds as your successes. (x-b choose 2)*(x-r choose 0)/ (x choose 2) = .5 Now, rewrite it (x-b)*(x-b-1)/(x)/(x-1) = .5 After multiplying out and completing the square you arrive at x = +/- (2b^2 + 1/4)^.5 +2b+.5 Only the positive is relevant here and of course x= r+b so now red as a function of black is r= (2b^2+1/4)^.5 + b+.5 Solve for values of black that yield perfect squares under the radical and you arrive at black = 1, red = 3 and b= 6, r=15

4. Thanks Soroban. Thanks Bilbo.

Bilbo I went through your method and got your final equation (though I didn't complete the square?). Did you then solve the equation by trial and error?

also do I need a perfect square under the radical? with the solutions inserted the evaluated radical does not give a whole number but sums to one with the terms b and 0.5

5. Originally Posted by factfinder
Thanks Soroban. Thanks Bilbo.

Bilbo I went through your method and got your final equation (though I didn't complete the square?). Did you then solve the equation by trial and error?

also do I need a perfect square under the radical? with the solutions inserted the evaluated radical does not give a whole number but sums to one with the terms b and 0.5
Well, you need to complete the square so that you can isolated X as the original quadratic is not a squared binomial. I then plugged in whole numbers for B (the domain for this specific problem is obviously restricted to whole numbers) starting from zero until I hit answers that yielded whole numbers for R (the range for this problem is restricted to whole numbers as well). In hindsight, I probably should have just plotted the function on a graphing calculator and searched for plausible results.
With regard to your second question, I do believe that a perfect square is anything that yields a rational number as a root. Therefore, I don't think that a perfect square must be an integer. I may be wrong in my use of the term "perfect square" though.

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