# Probability of Success on Multiple Unordered Trials

Here I will look at some events that do not have a specific order.

Example 1

### WithReplacement

You roll one die twice. What is the probability that you get one 5 and one 6?

There are two possible outcomes that can give you this result. One is first a 5 and then a 6. The second is to first get a 6 and then a 5. You first calculate the two probabilities separately and then add them together:

Using the same procedure, you calculate the second probability:

Now you have to put the two parts together by adding them together:

The probability of getting one 5 and one 6 when rolling one die twice is $\frac{1}{18}$.

When you calculate probability where one event occurs first, and then another event after, you multiply the two probabilities with each other (like when you calculated ). When you don’t care about the order, (as in the example above), you add the probabilities of the different outcomes together.

Example 2

### WithoutReplacement

You’re at a birthday party, and there are four bowls of snacks: Oreos, Doritos, Milk Duds and peanuts. You close your eyes and grab a bowl completely at random. Then you grab another bowl. What’s the probability that the two bowls you have drawn are the bowls with Oreos and Doritos?

There are two possible outcomes that can give us this result. The first is when you first get the bowl of Oreos and then the bowl of Doritos. The second is when you first get the bowl of Doritos and then the bowl of Oreos. You calculate the two probabilities separately:

Using the same procedure, you calculate the second probability:

Now you have to put the two parts together. The solution is to add the two probabilities together:

The probability of grabbing the bowls with Oreos and Doritos is $\frac{1}{6}$.

Notice that this probability is twice as high as the probabilities of grabbing the bowls in a given order such as first Oreos, then Doritos