﻿ Finding "and" probabilities with the multiplication rule - MathBootCamps

# Finding “and” probabilities with the multiplication rule

The probability of an “and” event, sometimes described as the intersection of two events, can be found using the multiplication rule for probability. In this guide, we will look at this formula and how to use it.

## The formula You can think of this in the following way: to find the “and” probability, find the probability of the first event and multiply it by the probability of the second event given the first. Let’s try it out with an example!

## Examples

### Example

A bag contains 25 blue and 15 green marbles. Two marbles are selected, one at a time, without replacement. What is the probability both are blue?

To answer this, you must first recognize that “both are blue” is the same as “the first is blue AND the second is blue”. Once you see this is an “and” probability, you can then apply the formula. $\text{P(both blue)} = \text{P(first is blue and second is blue)}$

To use the formula, think “probability of the first event times probability of second given the first”. This would give you the following: $\text{P(first is blue and second is blue)} = \text{P(first is blue)}\text{P(second is blue}|\text{first is blue)}$

Now we can find each probability. To find the probability the first marble is blue, notice that there are a total of 25 + 15 = 40 marbles and 25 are blue. Therefore: $\text{P(first is blue)} = \dfrac{25}{40}$

The next probability, P(second is blue | first is blue), is conditional. So, you must assume that the first marble was already selected and was blue. Since there were 40 marbles originally, there will only be 39 now. And, since there were 25 blue originally, there will only be 24 now. $\text{P(second is blue}|\text{first is blue)} = \dfrac{24}{39}$

Plugging these into the formula: $\text{P(first is blue and second is blue)} = \text{P(first is blue)}\text{P(second is blue}|\text{first is blue)}$ $= \left(\dfrac{25}{40}\right)\left(\dfrac{24}{39}\right)$ $\approx \boxed{0.3846}$

The key to this problem was the idea of “without replacement”. This meant that once the marble is selected, it is not put back. This changes the probability that another will be selected since there is 1 marble now gone.

Let’s look at another example that uses the same situation.

### Example

A bag contains 25 blue and 15 green marbles. Two marbles are selected, one at a time, without replacement. What is the probability the first is green and the second is blue?

In this example, it is more clear that we have an “and” probability, since it is stated right there in the question! So, we can immediately apply the formula. Again – think “probability of the first times probability of the second given the first”. $\text{P(first green and second blue)} = \text{P(first green)}\text{P(second blue}|\text{first green)}$ $= \left(\dfrac{15}{40}\right)\left(\dfrac{25}{39}\right)$ $\approx \boxed{0.2404}$

You may notice that we didn’t down each time here. Well, at the start, there are 40 total marbles and 15 are green, so the first probability is 15/40. Then, that marble is removed, so there are 39 remaining. But, since it was green, there are still 25 blue. This gives the conditional probability as 25/39.

## Conclusion

This is just the basics of finding probabilities with the multiplication rule. There are also cases where we have independent events (such as selecting WITH replacement) where the formula changes a bit and cases where using the complement may be helpful. 