Probability of Success on Multiple Ordered Trials

To calculate this probability, you should split the question into two parts, the first and the second roll.

First roll:

Here you look at the probability of getting a 5 when a die is rolled. You already know that it is $P(5)=\frac{1}{6}$.

Second roll:

Here you look at the probability of getting a 6 when a die is rolled. You don’t consider the previous throw, because the previous outcome of rolling the die doesn’t affect the next outcome at all! You know that the probability of getting a 6 when throwing a die is $P(6)=\frac{1}{6}$.

Now you have to put the two parts together. The way you do this is to multiply the two probabilities together:

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

The probability of first getting a 5, then a 6, is $\frac{1}{36}$ when you roll a single die twice.

You split the experiment into the first and second draw.

First draw:

Here you look at the probability of grabbing the bowl of Oreos. You know that $P(\text{Oreos})=\frac{1}{4}$.

Second draw:

Here you look at the probability of grabbing the bowl of Doritos. There is just one bowl that contains Doritos, so the number of favorable outcomes is also equal to 1. After already grabbing the bowl with Oreos, there are only three bowls left at the table. The number of possible outcomes is therefore equal to 3, making $P(\text{Doritos})=\frac{1}{3}$.

Now you have to put the two parts together. The solution is to multiply the two probabilities with each other:

The probability of first grabbing the bowl with Oreos and then the bowl with Doritos is $\frac{1}{12}$.