The Sock Drawer: Probability and Statistics Problem

A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2. 

a) How small can the number of socks in the drawer be?

b) How small if the number of black socks is even?

SOLUTION:

Let r red and b black socks. The probability of the first sock’s being red is \frac{r}{r+b}; and if the first sock is red, the probability of the second’s being red now that a red has been removed is \frac{r-1}{r+b-1}. Then we required the probability that both are red to be \frac{1}{2}, or

\frac{r}{r+b}\times \frac{\:r-1}{r+b-1}=\frac{1}{2}

Notice that

\frac{r}{r+b}>\frac{\:r-1}{r+b-1}

Therefore, we can create the inequalities

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

Taking the square roots, we have, for r>1.

\frac{r}{r+b}>\frac{1}{\sqrt{2}}>\frac{\:r-1}{r+b-1}

Simplifying, we have

\left(\sqrt{2}+1\right)b+1>r>\left(\sqrt{2}+1\right)b

So we can now easily plug in values for b, then solve for r.

When b=1, r must be between 2.414 and 3.414, and so r=3. For r=3,\:b=1,

P\left(2\:red\:socks\right)=\frac{3}{4}\cdot \frac{2}{3}=\frac{1}{2}

Therefore,\:the\:smallest\:number\:of\:socks\:is\:4..

Using the same inequality, we can substitute even values for b starting from 2, then solve for the value of r. After, check if the probability is 1/2. Refer to the table below.

 Investigation on the values of r and b

So, when b is 6, r is 15 and the probability of getting 2 red socks is 1/2. This condition satisfies the problem. Therefore,\:21\:is\:the\:smallest\:number\:when\:b\:is\:even..

 

 

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