Question

A drawer contains n white and n black socks. Each white sock has a unique design, and each black sock has a unique design. Two socks are selected at random from the drawer. Every way of selecting the two socks is equally likely, and the order in which the socks are selected does not matter.
(a) How many ways are there to select the two socks?
(b) How many ways of selecting the socks result in two socks of the same color being chosen?
(c)What is the probability that a randomly chosen pair of socks are the same color?

Answer 1

Answer:

a) 2n(2n-1)

b) 2n(n-1)

c) $\displaystyle\frac{2n(n-1)}{2n(2n-1)}=\frac{n-1}{2n-1}$

Step-by-step explanation:

There are in total n + n = 2n socks.

(a) How many ways are there to select the two socks?

The first sock can be selected in 2n different ways. The second sock can be selected in 2n-1 ways.

Since the order in which the socks are selected does not matter, by the Fundamental Principle of Counting there are

2n(2n-1)  

different ways of selecting two socks.

(b) How many ways of selecting the socks result in two socks of the same color being chosen?

There are n(n-1) different ways of selecting two white socks and n(n-1) ways of selecting two black socks, so there are

n(n-1) + n(n-1) = 2n(n-1)  

ways of selecting two socks of the same color.

(c)What is the probability that a randomly chosen pair of socks are the same color?

If every way of selecting the two socks is equally likely, the probability is

$\displaystyle\frac{2n(n-1)}{2n(2n-1)}=\frac{n-1}{2n-1}$