Advanced Math Topics 5.2 Adding Probabilities of Events

Mutually exclusive events-events that cannot happen at the same time Example: Drawing an ace and an 8 from a deck of cards are mutually exclusive events If A and B are mutually exclusive… p(A or B) = p(A) + p(B) Example: The probability of rolling a 2 or 3 on a die is1/6 + 1/6 = 2/6 = 1/3 NOTES

Complementary events-events that add to a probability of 1 P(A)P(complement) 77%23% /118/ %54.7% 100 – 77 = = 11/11 – 3/11 = 100 – 45.3 =

Addition Rule: If A and B are any events… p(A or B) = p(A) + p(B) – p(A and B) Example: p(red card or a Jack) =p(red) + p(Jack) -p(red jack) = 26/52 +4/52 -2/52 =28/52 = 14/26 = 7/13

These two formulas are actually the same formula. If A and B are mutually exclusive… p(A or B) = p(A) + p(B) Reason: The only difference is that in the first formula, we are not subtracting the probability of both events happening at the same time. NOTES Addition Rule: If A and B are any events… p(A or B) = p(A) + p(B) – p(A and B) We could add this into the formula, but since the first formula specifies that the events are mutually exclusive, the events cannot happen at the same time. Thus, subtracting p(A and B) would always be equal to subtracting 0.

2) According to the American Heart Association, 31% of people have high blood pressure, 40% of people are overweight, and 16% of people have high blood pressure and are overweight. If a person is randomly selected, what is the probability that this individual has high blood pressure or is overweight. From the HW (P. 247) p(A or B) = p(A) + p(B) – p(A and B) = 31% + 40% - 16% = 55%

7) 86% of the families in Monticello have either a dog or a cat. 64% have a dog and 33% have a cat. If a family is randomly selected, what is the probability that the family has both a dog and a cat? p(A or B) = p(A) + p(B) – p(A and B) 86% = 64% + 33% – x 86% = 97% – x x = 11%

HW P. 247 #1-11 Challenge Question if you choose to take it!