Probability of Success on Multiple Unordered Trials

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

Example 1

With Replacement

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 that you first roll a 5, and then a 6. The other is that you first get a 6, and then a 5. You first calculate the two probabilities separately and then add them together:

P (first 5, then 6) = \frac{1}{36}

Using the same procedure, you calculate the second probability:

\begin{array}{l} P (first 6, then 5) & = P (6) \cdot P (5) \\ = \frac{1}{6} \cdot \frac{1}{6} \\ = \frac{1 \cdot 1}{6 \cdot 6} \\ = \frac{1}{36} \end{array}

P (first 6, then 5) = P (6) \cdot P (5) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1 \cdot 1}{6 \cdot 6} = \frac{1}{36}

Now you have to put the two parts together using addition:

\begin{array}{l} P (5 and 6) & = P (first 5, then 6) \\ + P (first 6, then 5) \\ = \frac{1}{36} + \frac{1}{36} \\ = \frac{2}{36} \\ = \frac{1}{18} \end{array}

\begin{array}{l} P (5 and 6) & = P (first 5, then 6) + P (first 6, then 5) \\ = \frac{1}{36} + \frac{1}{36} \\ = \frac{2}{36} \\ = \frac{1}{18} \end{array}

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 that, you multiply the two probabilities together (like when you calculated

P (first 5, then 6)

). When you don’t care about the order, (as in the example above), you instead add the probabilities of the different outcomes together.

Example 2

Without Replacement

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:

\begin{array}{l} P (first Oreos, then Doritos) & = \frac{1}{4} \cdot \frac{1}{3} \\ = \frac{1 \cdot 1}{4 \cdot 3} \\ = \frac{1}{12} \end{array}

Using the same procedure, you calculate the second probability:

\begin{array}{l} P (first Doritos, then Oreos) & = \frac{1}{4} \cdot \frac{1}{3} \\ = \frac{1 \cdot 1}{4 \cdot 3} \\ = \frac{1}{12} \end{array}

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

\begin{array}{l} P (Oreos and Doritos) \\ = P (Oreos, Doritos) \\ + P (Doritos, Oreos) \\ = \frac{1}{12} + \frac{1}{12} \\ = \frac{2}{12} \\ = \frac{1}{6} \end{array}

\begin{array}{l} P (Oreos and Doritos) & = P (Oreos, Doritos) + P (Doritos, Oreos) \\ = \frac{1}{12} + \frac{1}{12} \\ = \frac{2}{12} \\ = \frac{1}{6} \end{array}

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.

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Probability of Success on Multiple Ordered Trials

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