**less than $\textbf6?$**

The problem you just did illustrates that probability is, in its basic essence, counting.

$\frac{ \text{outcomes that fit our condition} }{ \text{total number of outcomes overall} } = \frac{3}{8} .$

You can think of the fractions from probability as expressing a pair of counting problems: how many ways can we get what we want **vs.** how many outcomes there are total. The fundamental theorems of probability can also be built from this basic idea, and we'll do so in this course. For now, let's try solving a few more problems by just counting; it can be more powerful than you think!

Suppose you throw a pair of standard six-sided dice, one red and one blue.

Two ways to get a sum of $7$ are shown below:

Including the two examples above, how many ways are there to get a sum of $7$ on the pair of dice?

Now, if we ask,

"What is the probability of throwing a sum of $7$ on the dice we've been using?"

we can combine the facts we just worked out.

Our desired outcome is a sum of $7:$ there are $\textbf {6}$ ways to get this sum. The total number of outcomes possible is $\textbf {36}.$

So the probability we want is $\frac{6}{36} = \frac{1}{6} .$

Note it doesn't matter if we colored the dice or not! We colored them red and blue to emphasize that (for example) $3-4$ is considered different than $4-3.$ (Even when the dice *look* the same, they are physically different.)