SECTION 4.5
MULTIPLICATION RULE :
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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(1st and 2nd) = P(1st) * P(2nd |1st)
P(1st and 2nd) = 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(1st ∩ 2nd ∩ 3rd ∩ 4th ∩ 5th ∩ ….) 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)(798/806)…
25
P(25 people) = (798/806) = 0.779