1. QUANTITATIVE METHODS 1
SAMIR K. SRIVASTAVA

2. Probability Theory Basic Concepts
Uncertainty is part and parcel of life
Weather
Stock Market
Product Quality
A need arises to quantify the chances to facilitate better decision making.
Probability Theory provides us ways and means to attain the formal and precise expressions for uncertainties involved in different situations.
Probability: Quantitative measure of chance.
0.0 Absolutely no chance
1.0 Absolute certainty
4/22/2019
Quantitative Methods 1

3. Methods of Assigning Probabilities
Classical Approach
All elementary events are equally likely
With Replacement
Without Replacement
Relative Frequency Approach
Take a sample for the population
Take relative frequency of each event at an approximation of probability
Larger the sample, better the approximation. (Law of Large Numbers)
4/22/2019
Quantitative Methods 1

4. Methods of Assigning Probabilities
Coin Tossing P(H), P(T)
1.0
0.0
0.5
Subjective Approach
Based on past experience, judgement, belief.
Used when sampling is not possible (one time event)
4/22/2019
Quantitative Methods 1

5. Permutations How to list all possible outcomes of an experiment.
For a given experiment, count (enumerate) the no. of outcomes favorable to an event.
PERMUTATION is an arrangement of a given set of objects in a particular order.
a,b,c a,c,b b,a,c b,c,a c,a,b c,b,a
n objects can be arranged in n! different ways
P(n) = n(n-1)(n-2)…2.1
4/22/2019
Quantitative Methods 1

6. Permutations
Out of n object, r are selected and then arranged in some specific order. In how many different ways can this be done?
Select two letter out of a,b,c,d and arrange them in any order
a,b b,a a,c c,a a,c d,a b,c c,b b,d d,b c,d d,c
How many choices are available for the first letter? For the second letter?
P(n,r) = n(n-1)(n-2)…(n-r+1)
P(n,r)= n! / (n-r)!
4/22/2019
Quantitative Methods 1

7. Example
License plate numbers for cars have three alphabets followed by three digits, which can be intermixed. How many distinct license numbers are possible if no alphabet can be used twice on the same plate? What is the total number without this restriction?
possibilities for 3 alphabets : P(26,3) = 26!/23! = 15,600
Total possible numbers: X 1000 = 15,600,000
Without the restriction : 26 X 26 X 26 X 1000 = 17,576,000
4/22/2019
Quantitative Methods 1

8. Combination
In how many ways r objects can be selected from a set of n?
Ordering of objects is not important.
Select two letters out of a,b,c,d,e
a,b a,c a,d a,e b,c b,d b,e c,d c,e d,e
“n choose r” C(n,r) = P(n,r)/r! = n!/{(n-r)!(r)!}
nCr, (nr)
4/22/2019
Quantitative Methods 1

9. Combination
Example: Suppose that five positions of judges are to be filled in a certain state. Names of ten men and five women have been submitted to the president for these appointments. If the president decides that three men and two women should be appointed, in how many different can this be done?
Choose 3 men out of 10 : (103) = 120
Choose 2 women out of 5 : (52) = 6
Total number of ways : 120 X 6 = 720
4/22/2019
Quantitative Methods 1

10. Probability Theory Some terminology Experiment Tossing a die
Event (Experimental outcome) 1,2,3,4,5,6
Elementary Event 1,2,3,4,5,6
Composite Event < 3, even, odd
Composite event included more than one elementary outcomes.
Sample Space: A complete or exhaustive listing (set) of elementary events. {1,2,3,4,5,6}
All elements in the sample space are mutually exclusive and collectively exhaustive. (Composite events need not be mutually exclusive).
Assumption: All elements in the sample space are equally likely to occur
What is the chance of each elementary event occurring? Clearly 1 in 6 or 1/6 =
Can we now determine probabilities of composite events? How?
4/22/2019
Quantitative Methods 1

11. Composite Events Event A: outcome of the toss is an even number.
Which elementary outcomes are favorable to event A? 2,4,6.
Since each of these has a 1 in 6 chance, P(A) = 3/6 = 0.5
Event B: outcome is divisible by 3
Favorable outcomes: 3, 6. P(B) = 2/6 =
Complementary Event: The event in question does not occur.
Complement of event A is designated A’.
What is P(A’) ? Clearly 1-P(A). Why?
Sum of probabilities of events that are mutually exclusive and collectively exhaustive is always 1.0.
Odds: ratio of favorable to unfavorable outcomes.
What are the odds that the outcome is divisible by 3? 2:4
If the odds of an event are a:b, then its probability is a/(a+b).
4/22/2019
Quantitative Methods 1

12. Joint or Non-exclusive Events
Two events are non-exclusive if they have one or more elementrary outcomes common between them, i.e. they can occur together.
A: outcome is even.
B: outcome is divisible by 3.
Are they exclusive or non-exclusive? Why?
A Venn Diagram is useful 1 2 3 4 5 6
Sample space B A
Another example: Draw one card from a deck of 52 cards.
Event A: A Heart is drawn
Event B: A King is drawn
Are they exclusive?
A Black King is drawn
4/22/2019
Quantitative Methods 1

13. Probability of Union of Events
P(A or B) or P(AB)
If A and B are mutually exclusive events, P(A or B) = P(A)+P(B)
A – Outcome of die toss is less then 3
B – Outcome is divisible by 3 1 2 3 4 5 6
4/22/2019
Quantitative Methods 1

14. Probability of Union of Events
P(A or B) = 2/6+2/6 = 4/6 = 0.667
What if they are non-exclusive?
A – Even number
B – Divisible by 3
P(AB) = P(A)+P(B)-P(AB) General Rule of addition 1 2 3 4 5 6 A B
P(AB) = 3/6+2/6-1/6 = 4/6
4/22/2019
Quantitative Methods 1

15. Independent Events Draw a card from a deck. Put it aside.
Now draw another card.
What is the probability that first card is a King?
What is the probability that second card is a King?
Does it depend of the first draw? Why?
What if the first card is put back in the deck before the second draw? Does it still depend on the first draw? Why?
Two events are independent if the occurrence or non-occurrence of one event has no effect on the probability of the other event.
They are dependent when the occurrence of one event does affect the probability of the other event.
4/22/2019
Quantitative Methods 1

16. Conditional Probability
Event A : A King on the first draw.
Event B : A King on the second draw.
P(A) = 4/52
What about of P(B)? It depends on whether first draw was a King or not.
If A occurred, probability of B = 3/51
If A did not occur, probability of B = 4/51.
Notice that there is a condition attached to the probability in each case.
P(B/A) = 3/51
P(B/A’) = 4/51
4/22/2019
Quantitative Methods 1

17. Conditional Probability
Why is the probability different in the two cases? Dependence!
Test of Independence
Events A and B are independent if P(A/B) = P(A) and P(B/A) = P(B).
Here P(A) and P(B) are unconditional probabilities of the two events.
What is the unconditional (simple) probability of drawing a King on the second draw?
4/22/2019
Quantitative Methods 1

18. Joint Probability
For two events A and B, what is the probability that both events take place. P(A and B) or P(AB)
If A and B are independent events, P(A and B) = P(A).P(B)
What if they are dependent?
View this as a two step process:
A occurs.
Given that A has occurred, B occurs.
P(A and B) = P(A). P(B/A).
4/22/2019
Quantitative Methods 1

19. Joint Probability A die toss example:
A : outcome is even
B : outcome is a multiple of 3
P(A) = 3/6, P(B) = 2/6, P(B/A) = ?
New Sample Space 1 2 3 4 5 6 A B
P(A and B) = (3/6).(1/3) = 1/6
1/3
P(B/A) = P(A and B) / P(A)
4/22/2019
Quantitative Methods 1

20. Total Probability Formula
Consider dependent events A and B.
P(A) = P(AB) + P(A B)
= P(A/B)P(B) + P(A/B)P(B)
Consider a mutually exclusive and collectively exhaustive set of events: B1, B2, B3,…Bn
Let A be some event in the same sample space.
P(A) = P(AB1)+P(AB2)…..+P(ABn)
= P(A/B1)P(B1)+P(A/B2)P(B2)….+P(A/Bn)P(Bn) B1 B4 A B2 B3
4/22/2019
Quantitative Methods 1

21. Unconditional (Marginal) Probability of B
Revisit the Card experiment
A: King on first draw
B: King on second draw
We know the following probabilities:
P(A) = 4/52 P(A’) = 1-P(A) = 48/52
P(B/A) = 3/51 P(B/A’) = 4/51
Given these, can we compute P(B)?
4/22/2019
Quantitative Methods 1

22. Unconditional (Marginal) Probability of B
P(A and B) = P(A).P(B/A) = (4/52)(3/51)
P(A’ and B) = P(A’).P(B/A’) = (48/52)(4/51)
These two are mutually exclusive events
They include all possible outcomes favorable to event B.
P(B) = P(A and B) + P(A’ and B)
P(B) = P(A).P(B/A) + P(A’).P(B/A’)
= =
It is exactly the same as 4/52 = Why?
4/22/2019
Quantitative Methods 1

23. Bayes’ Theorem Revisit the Card drawing experiment again.
Event A : King on first draw
Event B : King on second draw
P(A), P(B/A) are easily determined. P(A) = 4/52, P(B/A) = 3/51.
Suppose we are interested in P(A/B)!!! How to find it?
Bayes’ theorem comes to our rescue.
Recall the joint probability formula
P(A and B) = P(A).P(B/A)
Reversing the role of events A and B, the following must also be true:
P(A and B) = P(B).P(A/B)
4/22/2019
Quantitative Methods 1

24. Bayes’ Theorem Combining the two, we get
P(A).P(B/A) = P(B).P(A/B)
Or P(A/B) = [P(A).P(B/A)]/P(B) Baye’s Formula
Let us substitute P(B) = P(A)P(B/A) + P(A’)P(B/A’)
P(A/B)
= [P(A).P(B/A)] / [P(A)P(B/A) + P(A’)P(B/A’)]
4/22/2019
Quantitative Methods 1

25. Another version of Bayes’ Formula
Consider a two step experiment.
The outcome of first step may be any one of the events A1, A2, A3,… An. These are mutually exclusive and exhaustive set of events for step 1.
Event B is a possible outcome of step 2 of the experiment.
P(Ai) and P(B/Ai) are known.
Can we find P(Ai/B) for any i.
Here P(B) = i P(Ai).P(B/Ai)
P(Ai/B) = P(Ai). P(B/Ai) /[i P(Ai).P(B/Ai)]
4/22/2019
Quantitative Methods 1

26. Another version of Bayes’ Formula
Need not be a two-step experiment.
Toss Die
A1 – 1 A2 – 2,3,4 A3 – 5,6
B – Multiple of 3
P(A1) = 1/6, P(A2) = 3/6, P(A3) = 2/6
P(B/A1) = 0, P(B/A2) = 1/3, P(B/A3) = ½
P(B) = 0.1/6 + 1/3 . 3/6 + ½ . 2/6 = 2/6
P(A3/B) = P(B/A3).P(A3)/P(B) = ½ . 1/3 / (1/3) = ½
4/22/2019
Quantitative Methods 1

27. Thank You !
4/22/2019
Quantitative Methods 1