Standard #: MA.912.DP.4.7

• Instruction includes connecting set operations (MA.912.LT.5.4) and exploring relationships using Venn Diagrams (MA.912.LT.5.5) when working to define the addition rule for probability: P(A or B) = P(A) + P(B) − P(A and B).
  
  
  
  • This can also be described as P(A ∪ B) = P(A) + P(B) − P(A ∩ B). 

• For example, when rolling a die, the probability for rolling a 4 or an even number can be calculated using the additional rule for probability, and expressed visually with Venn Diagrams. 
  • P(roll 4) = $\frac{\text{1}}{\text{6}}$ , P(roll even) = $\frac{\text{3}}{\text{6}}$ , and P(roll 4 and even) = $\frac{\text{1}}{\text{6}}$
    So, P(roll 4 or even) = $\frac{\text{1}}{\text{6}}$ + $\frac{\text{3}}{\text{6}}$ − $\frac{\text{1}}{\text{6}}$ = $\frac{\text{3}}{\text{6}}$ or $\frac{\text{1}}{\text{2}}$ .

• Emphasize that the additional rule for probability considers the probability or A or B to mean either A or B or both. 
• When developing the additional rule for probability, emphasize use of examples where the elements in each set may be easily counted. This will allow students to see that the elements in the intersection are counted twice, and therefore must be subtracted out once to avoid this double counting. 
  • For example, 10 students were surveyed about how they get to school each day. Aiden walks, Becky takes the bus, Charquanza takes the bus, Dennis rides a bike or walks, Elvis walks, Finnick walks, Grayson rides a bike or walks, Holly rides the bus, Isabel rides a bike, and Jayden rides a bike. If a student is chosen at random from this group, find the probability that they either ride a bike or walk to school. 
    • P(bike or walk) = P(bike) + P(walk) − P(bike and walk) 
    • P(bike or walk) = $\frac{\text{4}}{\text{10}}$ + $\frac{\text{5}}{\text{10}}$ − $\frac{\text{2}}{\text{10}}$ 
    • P(bike or walk) = $\frac{\text{7}}{\text{10}}$ 
  • Visually, we can show how Dennis and Grayson are double counted using a Venn Diagram, and therefore $\frac{\text{2}}{\text{10}}$ must be subtracted. 

• Instruction includes leading students to a definition of mutually exclusive by presenting cases where the probability of two events occurring simultaneously is 0, or impossible. For example, when rolling a die, the events rolling a 1 and rolling an even number are mutually exclusive. 
  • P(A or B) = P(A) + P(B) − P(A and B) 
  • P(1 or even) = P(1) + P(even) − P(1 and even) 
  • P(1 or even) = $\frac{\text{1}}{\text{6}}$ + $\frac{\text{3}}{\text{6}}$ − 0 
  • P(1 or even) = $\frac{\text{4}}{\text{6}}$ or $\frac{\text{2}}{\text{3}}$ 
• In this case, as there is no intersection, meaning P(A and B) = 0, we can use the formula P(A or B) = P(A) + P(B).

Common Misconceptions or Errors

• Students may not understand mutually exclusive events. 
• Students may neglect to subtract the intersection when using the addition rule for probability.

Instructional Tasks

• Today there is a 55% chance of rain, a 20% chance of lightning, and a 15% chance of lightning and rain together. 
  • Part A. Are the two events “rain today” and “lightning today” mutually exclusive? Justify your answer. 
  • Part B. What is the chance that we will have rain or lightning today? 
  • Part C. Now suppose that today there is a 50% chance of rain, a 60% chance of rain or lightning, and a 15% chance of rain and lightning. What is the chance that we will have lightning today?

Instructional Items

• At Mom’s diner, everyone drinks coffee. Let C represent the event that a randomly-selected customer puts cream in their coffee. Let S represent the event that a randomly-selected customer puts sugar in their coffee. Suppose that after years of collecting data, Mom has estimated the following probabilities: 
  
  P(C) = 0.6P(S)
                 = 0.5P(C or S)
     = 0.7 
• Estimate P(C and S) and interpret this value in the context of the problem.