6.3 – General Probability Rules “The most important questions of life are, for the most part, really problems of probability.” Pierre-Simon Laplace (1749-1827)

Addition Rule for Disjoint Events If they have no outcomes in common, then P(A or B or C) = P(A) + P(B) + P(C) This rule may extend to any number of disjoint events

Addition Rule for Joint Events If events can occur simultaneously, then the probability of their union is less than the sum of their probabilities. The outcomes common to both are counted twice when we add probabilities, so we must subtract this overlap once. P(A or B) = P(A) + P(B) – P(A & B)

Ex: Probability of Promotion Deborah and Matthew are anxiously awaiting word on whether they have been made partners in their law firm. Deborah guesses that her probability of being made partner is 0.7 and that Matthew’s is 0.5. Deborah also guesses that the probability that both she and Matt are made partners is 0.3.

Promotion continued P(at least one is promoted) P(D or M) = P(D) + P(M) – P(D & M) 0.7 + 0.5 – 0.3 = 0.9 P(no one is promoted) 1 – P(D or M) 1 – 0.9 = 0.1

Make a Venn Diagram to answer these Promotion Continued: P(D M) = 0.3 P(D MC) = 0.4 P(DC M) = 0.2 P(DC MC) = 0. 1 Make a Venn Diagram to answer these

From an actuary exam… A survey of a group’s viewing habits over the last year revealed the following info: (i) 28% watched gymnastics (ii) 29% watched baseball (iii) 19% watched soccer (iv) 14% watched gymnastics & baseball (v) 12% watched baseball & soccer (vi) 10% watched gymnastics & soccer (vii) 8% watched all three sports What % watched none of the three sports? 52%

HW: pg. 440 #65-70

Conditional Probability Probability of an event given that another event already happened. P(A|B) = Prob of A given that B has happened P(A|B) = P(A B) P(B|A) = P(A B) P(B) P(A) Conditional probability always reduces the sample space

Consider a family with two children There are 4 possible boy/girl combinations P(at least one girl) P(two girls) P(two girls|one child is a girl) Older Younger G B 3/4 1/4 1/3 Sample space is {GG, GB, BG}

Consider a family with two children P(two girls|oldest child is a girl) P(exactly one boy|one child is a girl) P(exactly one boy|youngest child is a girl) 1/2 Sample space is {GG, GB} 2/3 Sample space is {GG, GB, BG} 1/2 Sample space is {GG, BG}

Consider a family with two children If one of the children is a boy, what is the probability that the other child is a boy? If the oldest child is a boy, what is the probability that the other child is a boy? Many intelligent people would say 50% for both. HW: pg. 446 #6.71-6.76 1/3 Sample space is {BG, GB, BB} 1/2 Sample space is {BG, BB}

Independent Events Two events A and B are independent if P(B|A) = P(B) For independent events: P(A and B) = P(A)P(B) Dependent event: P(A and B) = P(A)P(B|A) HW: pg. 451 #6.80-6.82