1. NA387 Lecture 5: Combinatorics, Conditional Probability
(Devore, Chapter 2,
Sections: 2.3 – 2.4)

2. Topics Counting Techniques, Combinatorics Conditional Probability
Equally Likely Counting N(A)/N
Permutations
Combinations
Conditional Probability

3. Probability Counting Techniques
Equally Likely Counting (# outcomes is small)
Product Rule for Ordered Pairs
Tree Diagrams
Permutations
Combinations

4. Simple Counting Techniques
When various outcomes are “equally likely”, then computing probabilities reduces to counting.
P(A) = N(A) / N
Let’s define event A as: The Probability of rolling a Seven (sum of outcomes in rolling 2 fair dice);
N= 36 outcomes in sample space
Outcomes that yield a sum of 7: (1,6) , (6,1), (2,5) , (5,2) , (3,4) , (4,3)
N(A) = 6
N = 36 So, P(A) = 6/36 = 1/6
Note: Here N is relatively small
However, N may be quite large or outcomes may not be “equally likely”, so we need additional counting rules.

5. Product Rule for Ordered Pairs
Ordered Pairs: Select Two Objects
Object 1 – select n1 ways
Object 2 – select n2 ways
Total Number of pairs: n1*n2
Examples:
Roll Two Dice (6 outcomes per die)  total pairs: 36
Suppose every vehicle sold in a dealership has 1 engine and 1 transmission from the following options (outcomes):
Engine (Object A): 4-cyl, 6-cyl, 8-cyl
Transmission (Object B): Automatic, Manual
How may total pairs do you have?

6. Tree Diagrams- Example:
A production process consists of
machining (M) and finishing (F).
Suppose you have three machining
operations (M1, M2, M3). After
completion, products coming out of
M1 and M2 feed
finishing operations F1, F2, F3.
Whereas products coming out of M3
go to F4, F5, F6.
How many ordered pairs can we get? M1 M2 M3 F1 F2 F3 F4 F5 F6

7. General Product Rule
We may use the general product rule if we keep adding objects to form ordered collections of k-tuples (n1* n2*..nk ordered collections)
Ordered pair ~ 2-tuple
Three objects in collection ~ 3-tuple
Four objects in collection ~ 4-tuple, …
Suppose each of our engine/transmission collections have either a 5-year or 7-year warranty.
How many ordered collections are possible (ways in which one object of each may be selected)?
(hint: draw tree diagram, then compare with product rule)

8. With or Without Replacement?
An important question when identifying collections of objects is whether you select objects with or without replacement.
Prior examples assumed “with replacement” or “with repetition”, in other words, we assumed that we could pick as many objects as we wish without depleting the supply, say of the 4cyl transmissions…

9. Permutations
An ordered sequence of k objects are selected from a set of n distinct objects. (without replacement)
Total number of different permutations (arrangements) of n different objects = n!
Total number of permutations of n objects taken k at a time is:
(Note: m! ~ means m factorial) for instance: 4! = 4 x 3 x 2 x 1; 0! = 1

10. This is a permutation, since the beads will be in a row (order).
Example-Devore. A boy has 4 beads – red, white, blue, and yellow. How different ways can three of the beads be strung together in a row?
This is a permutation, since the beads will be in a row (order).
24 different ways
number selected
total

11. Ordered vs. Non-ordered
Another distinction to be made is whether we consider order in classifying outcomes.
Example: Suppose you have 3 circuit board locations and each board will require two components.
How does the number of possibilities change if the order of the location matters vs. if it does not matter?
Outcomes 1 2 3 A B 4 5 6
Board Locations

12. Combinations
A combination is an unordered subset of size k that can be selected from a set of n distinct objects.
Number of combinations of size k is much smaller than the number of permutations, because when order is disregarded, a number of permutations correspond to the same combination
Consider the prior example of 3 circuit board locations and two components per board. What is the number of Combinations?

13. Ex. A boy has 4 beads – red, white, blue, and yellow
Ex. A boy has 4 beads – red, white, blue, and yellow. How different ways can three of the beads be chosen to trade away?
This is a combination since they are chosen without regard to order.
total
number selected
4 different ways

14. Ex. Three balls are selected at random without replacement from the jar below. Find the probability that one ball is red and two are black.

15. Combinations: Example
Suppose a manufacturing bin of 50 parts contains 3 defective parts.
Samples of 6 parts are selected w/o replacement.
How many different ways are there to obtain samples that contain exactly 2 defects?
Is order important?
Note: This is (slightly) more challenging. Easier to break it into 5 steps as follows:

16. Defect Example Continued
Five Step Solution:
Find combinations of subsets that will yield exactly 2 defects (Event A)
Find combinations of subsets for the remaining 4 parts selected.
Find total possible subsets for exactly two defects is the product rule (n1 * n2) where n1 is number of collections of 2 defects and n2 is the number of collections of 4 non-defects.
Find total number of subsets of sample size (e.g., samples of 6)
P (Exactly 2 defects) = N(A) / N
Where N(A) = # subset combinations containing 2 defects
N = # subset combinations

17. A Further Complication
Suppose you want to determine the probability of finding at least 2 defects.
Hint:
At least 2  find 2 or 3 defects
Event A = exactly 2 defects, Event B = exactly 3 defects
P (A U B) = P (A) + P (B) – P (A B)
Note: P (A B) = 0 because mutually exclusive events
P (exactly 2) + P (Exactly 3)
Repeat the 5-step process again for exactly 3 defects. U U

18. Conditional Probability
Conditional probability is used when the outcome of an event may (or may not) change, given the outcome of a related event.
P(A|B) = Prob of A given B

19. Conditional Probability -- Interpretation
How should we interpret P(B | A)?
If all outcomes of an experiment are equally likely and there are n total outcomes, then:
P(A) = (number of A outcomes) / n
P(A B) = (number of outcomes in A and B)/n
So, P(B|A) = outcomes of A and B / outcomes of A
So, P(B | A) represents the relative frequency of event B among the trials that produce an outcome in event A. U

20. Conditional Probability Example
P(A|B) = P(A B) / P(B)
Example:
Event A: 90% of Car Model X Have Air Conditioning
Event B: 10% of Car Model X Have Sunroof
Event (A and B) : 8% of Car Model X Have Both
Given the Car has a Sunroof, what is the probability that it has air conditioning? U

21. Conditional Probability
Conditional probability is often derived from tree diagrams or contingency tables.
Suppose you manufacture 100 piston shafts.
Event A: feature A is not defective
Event B: feature B is not defective
P(A Not Def | B Def)?
P(A Not Def | B Not Def)?
P(A U B)?
P(B)?
Draw Tree Diagram.
P(A Not Def | B Def)? = 6/86 = 0.07
P(A Not Def | B Not Def)? 5/14 = 0.36

22. Solution to Defect Example (Slide 15)
Step 1: 3! / (2! * 1!) = 3
Step 2: 47! / (4! * 43!) = 178,365 ways
Step 3: 3 x 178,365 = 535,095  N(A)
Step 4: 50! / (6! * 44!) = 15,890,700  N
Step 5: P (exactly 2) = N(A) / N
P (exactly 2) = 535,095 / 15,890,700 = 0.034

23. Solutions Shaft Defects example : P(A Not Def | B Def)? = 6/86 = 0.07
P(A Not Def | B Not Def)? 5/14 = 0.36