SECTION 4 5 MULTIPLICATION RULE COMPLEMENTS AND CONDITIONAL

SECTION 4. 5 MULTIPLICATION RULE : COMPLEMENTS AND CONDITIONAL PROBABILITY

RECALL • P(A and B) • P(A∩B) • Involves the multiplication of the probability of event A and the probability of event B • If necessary, the probability of event B is adjusted because of the outcome of event A.

TYPES OF MULTIPLICATION RULES • Independent • P(A and B) = P(A∩B) = P(A) * P(B) • Dependent

DEPENDENT Two events are dependent if the occurrence affects the probability of the occurrence of the other. • Sampling without replacement is a dependent event. • P(A and B) = P(A∩B) = P(A) * P(B|A)

P(B|A) • Conditional Probability • Finds the Probability of event B after event A is removed.

EXAMPLE A • Find the probability when a day of the week is randomly selected, it is a Saturday and when a second different day of the week is randomly selected, it is a Monday.

EXAMPLE A • Find the probability when a day of the week is randomly selected, it is a Saturday and When a second different day of the week is randomly selected, it is a Monday. • P(Sat and Mon) = P(Sat) * P(Mon | Sat) • P(Sat and Mon) = (1 / 7) * (1 / 6) • P(Sat and Mon) = 1 / 42

EXAMPLE B • Acceptance sampling is a procedure that randomly selects items without replacement, and the entire batch is accepted if every item in the sample is okay. • Among 810 airport baggage scales, 102 are defective. If four of the scales are randomly selected and tested, what is the probability that the entire batch will be accepted?

EXAMPLE B • Independent or Dependent?

EXAMPLE B • Dependent • Set – up?

EXAMPLE B Set – up P(A∩B∩C∩D) = P(A)*P(B|A)*P(C|A∩B)*P(D|A∩B∩C) • A = sample #1 • B = sample #2 • C = sample #3 • D = sample #4

EXAMPLE B P(A) = total okay / total number • P(B|A) = total okay minus first pick / total number minus first pick • P(C|A∩B) = total okay minus first two picks / total number minus the first two picks • P(D|A∩B∩C) = total okay minus first three picks / total number minus the first three picks.

EXAMPLE B • P(A) = 708 / 810 • P(B|A) = 707 / 809 • P(C|A∩B) = 706 / 808 • P(D|A∩B∩C) = 705 / 807

EXAMPLE B • P(A∩B∩C∩D) = P(A) * P(B|A) * P(C|A∩B) * P(D|A∩B∩C) = (708 / 810 ) * (707 / 809) * (706 / 808) * (705 / 807) = 0. 583

CUMBERSOME CALCULATIONS • Sometimes the dependent calculations can become tedious so on certain problems we can treat our dependents and independents.

CUMBERSOME CALCULATIONS • TREATING DEPENDENT EVENTS AS INDEPENDENT • 5% GUIDELINE FOR CUMBERSOME CALCULATIONS • When calculations with sampling are very cumbersome and the sample size is no more than 5% of the size of the population, treat the selections as being independent.

EXAMPLE • Among respondents asked which is their favorite seat on a plane, 492 chose the window seat, 8 chose the middle seat, and 306 chose the aisle seat.

EXAMPLE 1 What is the probability of randomly selecting 1 of the surveyed people and getting one who did not choose the middle seat?

EXAMPLE 1 What is the probability of randomly selecting 1 of the surveyed people and getting one who did not choose the middle seat? P(one middle seat) = (492 + 306) / (492 + 8 + 306) P(one middle seat) = 798 / 806 = 0. 990

EXAMPLE 1 Now find me a second way to solve this problem What is the probability of randomly selecting 1 of the surveyed people and getting one who did not choose the middle seat?

EXAMPLE 1 Use Complements

EXAMPLE 1 What is the probability of randomly selecting 1 of the surveyed people and getting one who did not choose the middle seat? P(Aisle or Window) = 1 – P(middle) P(Aisle or Window) = 1 – (8 / 806) P(Aisle or Window) = 1 – 0. 0099255583 P(Aisle or Window) = 0. 99007

EXAMPLE 2 If 2 of the surveyed people are randomly selected without replacement, what is the probability that neither of them chose the middle seat?

EXAMPLE 2 If 2 of the surveyed people are randomly selected without replacement, what is the probability that neither of them chose the middle seat? P(1 st and 2 nd) = P(1 st) * P(2 nd |1 st) P(1 st and 2 nd) = 798/806 * 797/805 = 0. 980

EXAMPLE 3 If 25 different surveyed people are randomly selected without replacement, what is the probability that none of them chose the middle seat?

EXAMPLE 3 If 25 different surveyed people are randomly selected without replacement, what is the probability that none of them chose the middle seat? P(1 st ∩ 2 nd ∩ 3 rd ∩ 4 th ∩ 5 th ∩ …. ) and it is without replacement, so it is dependent.

EXAMPLE 3 Does it fall within the 5% cumbersome calc? 25 people / 1000 total people = 0. 025 = 2. 5% Thus we can use the 5% cumbersome calc.

EXAMPLE 3 If 25 different surveyed people are randomly selected without replacement, what is the probability that none of them chose the middle seat? P(25 people) = (798/806)… 25 P(25 people) = (798/806) = 0. 779