# Probability of getting exactly two balls of the same color

Suppose you have $5$ boxes, each of which contains a red ball, a blue ball, a green ball, a yellow ball and a white ball.

If you draw a ball at random from each box, what is the probability that there are exactly two balls of the same color?

Stuck a small amount with this question:

The probability of getting two balls of the same color is

$$P(\text{2 Same Color}) = 1 \cdot \frac{1}{5} \cdot \frac{4}{5} \cdot \frac{3}{5} \cdot \frac{2}{5} = \frac{24}{625}$$

Then just multiplying $P(\text{$2$Same Color})$ by $5$ as there are $5$ different colors of balls so you could get $2$ whites or $2$ reds etc,

That gives you

$$\frac{24}{625} \cdot 5 = \frac{24}{125}$$

Not sure what the answer to this is just looking for someone to tell me if I'm on the right track or completely wrong.

• Does " exactly two balls of the same color" includes $RRBBY$ as well? May 15 '18 at 17:45
• Explain where each factor comes from in $1\cdot \frac 15\cdot \frac 45\cdot \frac 35\cdot \frac 25.$ Doesn't this account for 5 different colors? What it doesn't account for is placement of the two balls of the same color. May 15 '18 at 17:57
• @callculus I dont believe so although it doesn't say in the question. May 15 '18 at 18:02
• @DougM After you saying that I can see that is probably wrong. My thoughts were 1 as the first ball you pick has to be unique in color and then $\frac{1}{5}$ as the second ball has to be the same color as the first so $\frac{1}{5}$ and then $\frac{4}{5}$, $\frac{3}{5}$ and $\frac{2}{5}$ as the probability of not getting the same color again May 15 '18 at 18:04

Pick the color of the pair ($5$) then choose the two boxes that you draw that color from ($_5C_2 = 10$). Then, pick three colors other than the pair's color $(_4C_3 = 4)$ and choose the order that those colors are picked out of the other three boxes $(3! = 6)$.
That gives $1200$ ways to pick exactly one pair. Divide by $5^5 = 3125$ ways to pick any five balls, and you have your answer.