Conditional Probability

Today, we’re going to take a look at conditional probability. This type of probability is all about finding the odds of something happening given that something else has already occurred.

Let’s start with a refresher on basic probability. The probability of something happening can be expressed as the number of desired outcomes divided by the number of total possible outcomes:

\(P(A)=\frac{\text{Number of desired outcomes}}{\text{Total number of possible outcomes}}\)

Here’s an example of a basic probability problem. There are 10 marbles in a bag: 3 are blue, 3 are red, and 4 are orange. What are the odds of pulling out a red marble? To solve this, we’d place the number of desired outcomes on the top (in this case, it’s 3 because there are 3 red marbles) and the total number of possible outcomes on the bottom (this is 10 since there are 10 total marbles in the bag).

\(P(A)=\frac{\text{Number of desired outcomes}}{\text{Total number of possible outcomes}}=\frac{3\text{ red marbles}}{10\text{ total marbles}}\)

Seems pretty straightforward, right? Well, conditional probability adds a bit of a twist to this, as the objects or people in question often have more than one possible attribute.

Let’s look at another example problem with more than one possible attribute.

Out of 100 houses sold, 50 had a garage, 40 had a pool, and of these, 10 had both a garage and a pool, leaving 20 houses with neither. Given that a house had a garage, what are the odds that it also had a pool?

What do we do with all this information> If we draw it as a Venn diagram, we can make more sense of it:

We can now see the overlap between the houses sold with garages and the houses sold with pools, all within the larger set of houses sold. It’s easier to see now that of the 50 houses sold with a garage, 10 of them also had a pool. So the odds are \(\frac{10}{50}\), which can be expressed as the fraction \(\frac{1}{5}\), or 20%.

Fortunately, there’s a faster way to do this than by drawing a diagram every time you encounter a conditional probability question.

Conditional Probability Formula

There’s a handy formula we can use:

\(P(B|A)=\frac{P(B\text{ and }A)}{P(A)}\)

We read this formula as “the probability P of event B happening given that event A has happened is equal to the probability of events B and A both happening over the probability that event A has happened.”

In these types of problems, both probabilities on the right side of the equation are usually given or can be determined from a sentence or table. So in our sample problem, the \(P(B|A)\) is the 10 houses that were sold with both a garage and a swimming pool, out of the total 100 houses. This fraction can be written as \(\frac{1}{10}\).

The P(A) is the probability of a house being sold with a garage (event A), which is 50 houses out of the 100, or \(\frac{5}{10}\). Now that our formula is complete, we can divide \(\frac{1}{10}\) by \(\frac{5}{10}\) and we get \(\frac{1}{5}\), which is equal to 20%, just as we figured out from our diagram.

Let’s try one more with numbers that aren’t quite so neat.

In a small town, there are 216 men: 144 of them have brown hair and 56 of them have mustaches. 24 of the men have brown hair and mustaches. What are the odds that a man chosen at random has a mustache given that he has brown hair?

Let’s plug into our formula. The probability of event A (brown hair) and event B (mustache) is the 24 men with both, divided by the 216 total men.

The probability of event A is 144 divided by the 216 total men. We divide \(\frac{24}{216}\) by \(\frac{144}{216}\) using fraction division. The 216s can be canceled out, so we are left with \(\frac{24}{144}\). This fraction can be reduced to \(\frac{1}{6}\), or ≈16.67%.

All right, that’s all for this review. I hope this increased your odds of understanding conditional probability.

Thanks for watching, and happy studying!

Conditional Probability Practice Questions