Probability of Success on Multiple Ordered Trials
So far, we have looked at how to work with probabilities when you have rolled one die, or otherwise had one single trial. We called these scenarios experiments with one trial, because the experiment is performed once. Now, let’s look at with probabilities with multiple trials.
Example 1
Experiments with Replacement
You roll a die twice. What’s the probability of getting a 5 and then a 6?
To calculate this probability, you should split the question into two parts, the first and the second roll.
- First roll:
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Here you look at the probability of getting a 5 when a die is rolled. You already know that it is .
- Second roll:
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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 .
Now you have to put the two parts together. The way you do this is to multiply the two probabilities together:
The probability of first getting a 5, then a 6, is when you roll a single die twice.
Example 2
Experiments 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 at random. You then grab a different bowl, also at random. What’s the probability that the first bowl you grabbed contained Oreos? What’s the probability that the second bowl you grabbed contained Doritos?
You split the experiment into the first and second draw.
- First draw:
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Here you look at the probability of grabbing the bowl of Oreos. You know that .
- Second draw:
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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 .
Now you have to put the two parts together. The solution is to multiply the two probabilities with each other:
Example 1 is a problem with replacement. You can’t remove one side of the die for your second roll. All sides of the die are included in all die rolls. “With replacement” means that you have the same number of possible outcomes in the next trial as in the previous trial.
Example 2 is a problem without replacement. You can’t return the bowl of snacks that you’ve already grabbed. “Without replacement” means that the number of possible outcomes changes. Those are experiments where you don’t put back—or replace—what you drew last time.
