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# Common Core Standards: Math

# Math.CCSS.Math.Content.HSS-CP.B.6

**6. Find the conditional probability of A given B as the fraction of B's outcomes that also belong to A, and interpret the answer in terms of the model.**

Do you remember the last time you went out for pizza with a couple of friends? The three of you sat there laughing and eating away at the 8-slice, New York style, pepperoni pizza in front of you. You finished up your second slice and grabbed a sip of soda. At the same time, Billy finished his second slice. Realizing what was going on, Bob stuffed the his remaining second slice into his mouth.

Only two slices remained for the three of you. An all-out brawl was about to erupt. As the pacifist of the group, you grabbed a plastic knife and cut the 2 slices into three equal portions, narrowly averting the mozzarella melee. Always the resourceful one.

But what if you hadn't been so resourceful? Before you started eating, what is the probability you would have exactly three slices of pizza, *P*(*A*)? Once you've each had two slices, what is the probability you'd have exactly three slices, *P*(*A*|*B*)? (Here, event *B* is you each have had 2 slices, and there are only two left.)

Students should already recognize that these probabilities, *P*(*A*) and *P*(*A*|*B*) are different, but they may not necessarily understand why. They know how to calculate P(*A*|*B*) using the formula

Students should be able to identify why these two probabilities are not always equal. They should also understand that *P*(*A*|*B*) represents the outcomes remaining for *A* to occur once *B* has already occurred. This is the fraction of outcomes of *B* that also belongs to *A*.

Let's go back to the pizza problem. We can count all of the possible outcomes, and from them count the number that result in 3 slices for you.

You | Billy | Bob |

1 | 1 | 6 |

1 | 2 | 5 |

1 | 3 | 4 |

1 | 4 | 3 |

1 | 5 | 2 |

1 | 6 | 1 |

2 | 1 | 5 |

2 | 2 | 4 |

2 | 3 | 3 |

2 | 4 | 2 |

2 | 5 | 1 |

3 | 1 | 4 |

3 | 2 | 3 |

3 | 3 | 2 |

3 | 4 | 1 |

4 | 1 | 3 |

4 | 2 | 2 |

4 | 3 | 1 |

5 | 1 | 2 |

5 | 2 | 1 |

6 | 1 | 1 |

We see there are 4 out of 21 situations where you'll get exactly 3 slices. The probability of this occurring, call it event *A*, is ^{4}⁄_{21} ≈ 0.19. The probability changes when you are all staring each other down after having two slices apiece. We can count all the possible outcomes where each of you have had *at least *2 slices.

You | Billy | Bob |

2 | 2 | 4 |

2 | 3 | 3 |

2 | 4 | 2 |

3 | 2 | 3 |

3 | 3 | 2 |

4 | 2 | 2 |

This time, there are only 6 possible outcomes, and 2 of them result in you getting exactly 3 slices. Your chances have gone up! *P*(*A*|*B*) = ^{2}⁄_{6} ≈ 0.33. Students should recognize that we've counted the fraction of outcomes of event *B* (you've all had *at least* 2 slices) that *also *belong to *A* (you've had exactly 3 slices).

We want to connect this to what we already know. Using our formula for conditional probability, we know *P*(*A* and *B*) occurring is the probability you've each had at least 2 slices and you have had 3. From our original list, we narrow it down to 2 out of 21 outcomes.

You | Billy | Bob |

3 | 2 | 3 |

3 | 3 | 2 |

Meanwhile, *P*(*B*) is the number of outcomes in each of you all having at least 2 slices, which we already know is 6 out of 21 outcomes. Doing the math,

Finally, visualizing this idea for the students with a Venn diagram may help them understand more clearly.

Here, each dot represents a possible outcome. There are 6 dots in *B*, and there are 2 dots in *A* and *B*.