QUANTITATIVE METHODS 1

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

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
Methods of Assigning Probabilities

• Coin Tossing P(H), P(T)

• Subjective Approach
• Based on past experience, judgement, belief.
• Used when sampling is not possible (one time
  event)

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

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-r1)
• P(n,r) n! / (n-r)!

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 15600 X 1000
  15,600,000
• Without the restriction 26 X 26 X 26 X 1000
  17,576,000

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)

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

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 lt 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 0.1667.
• Can we now determine probabilities of composite
  events? How?

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 0.333.
• 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? 24
• If the odds of an event are ab, then its
  probability is a/(ab).

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

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
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

14
Probability of Union of Events
P(A or B) 2/62/6 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
P(A?B) 3/62/6-1/6 4/6
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.

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

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?

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).

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) ?

P(A and B) (3/6).(1/3) 1/6
1/3
P(B/A) P(A and B) / P(A)
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)

B4
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)?

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)

0.00450.0724 0.0769
It is exactly the same as 4/52 0.0769. Why?
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)

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)
  Bayes 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)

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)

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

27
Thank You !