standard_description: 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.
problem_activity: Problem 1
Each scenario on the left corresponds with a calculation on the right. Connect the scenarios and calculations and identify which features are most useful in connecting the representations:
|              Rolling once the number is an odd number or a number less than 3.              | $${P(A\space\mathrm{and}\space B) = \left({3\over6}\right)\left({1\over6}\right)}$$ |
|---------------------------------------------------------------------------------------------|-------------------------------------------------------------------------------------|
|                     Rolling once and the number is an odd number or 4.                      |         $${P(A\space\mathrm{or}\space B) = {3\over6}+{2\over6}-{1\over6}}$$         |
| Rolling twice. The first roll is an odd number and the second roll is a number less than 3. | $${P(A\space\mathrm{and}\space B) = \left({3\over6}\right)\left({2\over6}\right)}$$ |
|         Rolling twice. The first roll is an odd number and the second roll is a 4.          |              $${P(A\space\mathrm{or}\space B) = {3\over6}+{1\over6}}$$              |
Problem 2
Below is a spinner with four quadrants, each labeled 1–4. Each outcome is equally likely.
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Sam spins the spinner twice and doesn’t land on 4. What is the probability of this occurring?
Problem 3
Allison has put together a simple game. She put 6 cubes in a paper bag—3 yellow and 3 blue. Allison has determined that the rules are as follows:
Pull a cube from the bag and put it on the table.
Pull a second cube from the bag.
If the cubes are different colors, then Player A wins.
If the cubes are the same color, then Player B wins.
Is this game fair? How do you know?


aligns: no