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# Common Core Standards: Math

# Math.CCSS.Math.Content.HSS-CP.A.3

**3. Understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B.**

You're taking a weeklong vacation. You booked a quiet, lakeside cabin, nestled at the foot of the unadulterated mountains a few hours from home. Although you have a good idea of where you're going, you still lean on your GPS—you named her Geepas—to get you safely to your respite from the world.

As you get within a few miles of the cabin, Geepas takes you down a dingy road with a tree canopy that blocks out the sunlight. Soon, the pavement turns to dirt, and you arrive at a dead end. You know you're close, but you're lost in a foreign wilderness. Frantically, you search for a map, any sign of civilization, or a beam of sunlight that can point you in the right direction. Geepas has failed you.

This is how students are likely to feel the first time you teach them conditional probability. You will be their Geepas, leading them down a path to follow. If you aren't careful about how you present the ideas, you many leave them stranded.

**Conditional probability** is really about asking a simple question: now that event *A* *has* occurred, what is the probability that event *B will* occur? The Venn diagram in the image depicts this idea. Let's say we already know that event *A* has occurred. The possible outcomes for event *B* are those shaded in black. Likewise, the possible outcomes for event *C* are those shaded in green.

Now you can ask the students, "Once event *A* has already occurred, are the possible outcomes for event *D* those shaded in orange?" Many students may jump up with enthusiasm, yelling, "Yes! Orange!" Geepas led those students astray. The correct answer is that *there are no possible outcomes* for event *D* given that event *A* has already occurred. It's all right to lead them astray, so long as you can lead them back on the right track. Before the battery life runs out, preferably.

We want to break down what happened here. We already know that event *A* occurred. It happened. It's a fact. Instead of all possible outcomes being in the space *Ω*, all possible outcomes are in the blue space. Since *A* occurred, what is the probability that event *B* occurred? This is called *P*(*B*|*A*), read "the probability of *B* given *A*." We have fewer outcomes to choose from than before, now that *A* has occurred. The probability of *B* occurring should reflect that.

Now that you've drawn the distinction for the students, you can present the formula on how to calculate *P*(*B*|*A*) for the students:

If we look at this formula, we are essentially just renormalizing our total list of outcomes. This idea of renormalizing may not be clear to the students. It'll be helpful to explain it using the Venn diagram to show them that you are just dividing by the total space of event *A* because the probability of event *A* occurring is 1.00.

Let's use some numbers. Let's say we are told that *P*(*A*) = 0.70, *P*(*B*) = 0.30, and *P*(*A* and *B*) is 0.15. This is before event *A* or *B* has occurred. Now we are told event *A* has occurred. What is the probability that *B* will occur, *P*(*B*|*A*)?

It's *not* 0.30. Using our formula, we know that:

The probability that *B* will occur after *A* has occurred actually decreases. This means that the two events appear to be related in some way.

What is the probability of *A* occurring after *B* has occurred, *P*(*A*|*B*)?

The probability of *A* *increases* after *B* has occurred. Now it is clear that events A and B are not independent. Mathematically, we require *P*(*A*|*B*) ≠ *P*(*A*) or *P*(*B*|*A*) ≠ *P*(*B*) if events *A* and *B* are not independent.

How do we know if two events are independent? Let's say the probability of event *C* is 0.25 and *P*(*A* and *C*) = 0.175. Well, we just have to make sure that these equations are true.

Since *P*(*A*|*C*) = *P*(*A*) and *P*(*C*|*A*) = *P*(*C*), events *A* and *C* are independent.

We want to be careful to draw one last distinction. Let's look at the Venn diagram for events *A* and *D*. It's clear that *P*(*A* and *D*) = 0.00. Let's say *P*(*D*) = 0.02. Given our formula, *P*(*D*|*A*) = *P*(*A*|*D*) = 0.00. Note that these are not equal to *P*(*D*) or *P*(*A*). Events *A* and *D* are not independent. The Venn diagram can be a little misleading here. Students may have trouble with this idea, so be sure to steer them to the formula, not the diagram, to determine independence.

In general, this particular subject is murky. As Geepas, you can help the students find their way. Relating to them that Venn diagrams are useful visual tools is important. Here, it can be used to explain what conditional probability is. After that, you should break the students free of the Venn diagram, guiding the students to the formula to help them determine independence of events.