1. MULTIPLICATION RULE : COMPLEMENTS AND CONDITIONAL PROBABILITY
SECTION 4.5
MULTIPLICATION RULE : COMPLEMENTS AND CONDITIONAL PROBABILITY

2. 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.

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

4. 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)

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

6. 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.

7. 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

8. 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?

9. EXAMPLE B
Independent or Dependent?

10. EXAMPLE B
Dependent
Set – up?

11. 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

12. 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.

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

14. 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

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

16. 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.

17. 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.

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

19. 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) = ( ) / ( )
P(one middle seat) = 798 / 806 = 0.990

20. 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?

21. EXAMPLE 1
Use Complements

22. 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 –
P(Aisle or Window) =

23. 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?

24. 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(1st and 2nd) = P(1st) * P(2nd |1st)
P(1st and 2nd) = 798/806 * 797/805 = 0.980

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

26. 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(1st ∩ 2nd ∩ 3rd ∩ 4th ∩ 5th ∩ ….) and it is without replacement, so it is dependent.

27. EXAMPLE 3 Does it fall within the 5% cumbersome calc?
25 people / 1000 total people
= = 2.5%
Thus we can use the 5% cumbersome calc.

28. 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)(798/806)…
P(25 people) = (798/806)25 = 0.779