G10 Math Q4-Week 1- Mutually Exclusive.ppt

math 10

1. Mutually and
Non
Mutually
Exclusive

2. Theorems of Probability
There are 2 important theorems of
probability which are as follows:
2
The Addition Theorem and
The Multiplication Theorem

3. Addition theorem when events are Mutually
Exclusive
Definition: - It states that if 2 events A and B are mutually exclusive then the probability of the occurrence of either A or B is the sum
of the individual probability of A and B.
Symbolically

3
P(A or B) or P(A U B) = P(A) + P(B)
P(A or B or C) = P(A) + P(B) + P(C)
The theorem can be extended to three or more mutually
exclusive events. Thus,

4. Addition theorem when events are not Mutually
Exclusive (Overlapping or Intersection Events)
Definition: - It states that if 2 events A and B are not mutually exclusive then the probability of
the occurrence of either A or B is the sum of the individual probability of A and B minus the
probability of occurrence of both A and B.
Symbolically

4
P(A or B) or P(A U B) = P(A) + P(B) – P(A ∩ B)

5. Mutually Exclusive Events
Two events are mutually exclusive if
they cannot occur at the same time
(i.e., they have no outcomes in
common).
In the Venn Diagram above,
the probabilities of events A
and B are represented by
two disjoint sets (i.e., they
have no elements in
common).
Non-Mutually Exclusive Events
Two events are non-mutually exclusive if
they have one or more outcomes in
common.
In the Venn Diagram above, the
probabilities of events A and B
are represented by two
intersecting sets (i.e., they have
some elements in common).

6. The Addition Rule: Mutually Exclusive
P(A or B) = P(A) + P(B)
The Addition Rule: Non-mutually Exclusive
P(A or B) = P(A)+P(B) - P(A and B)
Probability of A and B
happening together
Probability of B
happening
Probability of A
happening
Probability of A or B
happening when and B are
not Mutually exclusive
Probability of either A or B happening

7. Multiplication theorem
Definition: States that if 2 events A and B are
independent, then the probability of the occurrence
of both of them (A & B) is the product of the individual
probability of A and B.
Symbolically,
Probability of happening of both the events:
P(A and B) or P(A ∩ B) = P(A) x P(B)
P(A, B and C) or P(A ∩ B ∩ C) = P(A) x P(B) x P(C)
Theorem can be extended to 3 or more independent events.
Thus,

8. How to calculate probability in case of
Dependent Events
Case Formula
1. Probability of occurrence of at least A or B
1.When events are mutually
2.When events are not mutually exclusive
2. Probability of occurrence of both A & B
3. Probability of occurrence of A & not B
4. Probability of occurrence of B & not A
5. Probability of non-occurrence of both A & B
6. Probability of non-occurrence of atleast A or B
P(A U B) = P(A) + P(B)
P(A U B) = P(A) + P(B) – P(A ∩ B)
P(A ∩ B) = P(A) + P(B) – P(A U B)
P(A ∩ B) = P(A) - P(A ∩ B)
P(A ∩ B) = P(B) - P(A ∩ B)
P(A ∩ B) = 1 - P(A U B)
P(A U B) = 1 - P(A ∩ B)
8

9. How to calculate probability in case of
Independent Events
Case Formula
1. Probability of occurrence of both A & B
2. Probability of non-occurrence of both A
& B
3. Probability of occurrence of A & not B
4. Probability of occurrence of B & not A
5. Probability of occurrence of atleast one
event
6. Probability of non-occurrence of atleast
one event
7. Probability of occurrence of only one
event
P(A ∩ B) = P(A) x P(B)
P(A ∩ B) = P(A) x P(B)
P(A ∩ B) = P(A) x P(B)
P(A ∩ B) = P(A) x P(B)
P(A U B) = 1 - P(A ∩ B) = 1 – [P(A) x P(B)]
P(A U B) = 1 - P(A ∩ B) = 1 – [P(A) x P(B)]

P(A ∩ B) + P(A ∩ B) = [P(A) x P(B)] +
[P(A) x P(B)]

10. Problem
An inspector of the Alaska Pipeline has the task of
comparing the reliability of 2 pumping stations.
Each station is susceptible to 2 kinds of failure: Pump
failure & leakage. When either (or both) occur, the
station must be shut down. The data at hand
indicate that the following probabilities prevail:
StationP(Pump failure)P(Leakage)P(Both)
10.07 0.10 0
20.09 0.12 0.06
Which station has the higher probability of being
shut down.
10

11. Solution
P(Pump failure or Leakage)
= P(Pump Failure) + P(Leakage Failure)
– P(Pump Failure ∩ Leakage Failure)
11
Station 1: = 0.07 + 0.10 – 0
= 0.17
Station 2: = 0.09 + 0.12 – 0.06
= 0.15
Thus, station 1 has the higher
probability of being shut down.

12. Probability Rules

13. Probabilities under conditions of
Statistical Independence
Statistically Independent Events: - The
occurrence of one event has no effect on the
probability of the occurrence of any other
event
Most managers who use probabilities are
concerned with 2 conditions.
1. The case where one event or another will occur.
2. The situation where 2 or more. Events will both occur.

14. There are 3 types of probabilities under
statistical independence.
Marginal
Joint
Conditional
Marginal/ Unconditional Probability:
- A single probability where only one event can take

place.
.
Joint probability:
- Probability of 2 or more events occurring together or in
succession.
Conditional probability:
- Probability that a second event (B) will occur if a first
event (A) has already happened

15. Example: Marginal Probability - Statistical Independence
A single probability where only one event
can take place.
Marginal Probability of an Event
P(A) = P(A)
Example 1: - On each individual toss of an biased or unfair
coin, P(H) = 0.90 & P(T) = 0.10. The outcomes of several
tosses of this coin are statistically independent events too,
even tough the coin is biased.
Example 2: - 50 students of a school drew lottery to see
which student would get a free trip to the Carnival at Goa.
Any one of the students can calculate his/ her chances of
winning as:
P(Winning) = 1/50 = 0.02

16. Example: Joint Probability - Statistical Independence
The probability of 2 or more independent events
occurring together or in succession is the product of their
marginal probabilities.
Joint Probability of 2 Independent Events
P(AB) = P(A) * P(B)
Example: - What is the probability of heads on 2
successive tosses?
P(H1H2) = P(H1) * P(H2)
= 0.5 * 0.5 = 0.25
The probability of heads on 2 successive tosses is
0.25, since the probability of any outcome is not
affected by any preceding outcome.

17. We can make the probabilities of events even more
explicit using a Probabilistic Tree.
1 Toss 2 Toss 3 Toss
H10.5 H1H2 0.25 H1H2H3 0.125
T10.5 H1T2 0.25 H1H2T3 0.125
T1H2 0.25 H1T2H3 0.125
T1T2 0.25 H1T2T3 0.125
T1H2H3 0.125
T1H2T3 0.125
T1T2H3 0.125
T1T2T3 0.125

18. Example: Conditional Probability - Statistical Independence
For statistically independent events, conditional
probability of event B given that event A has occurred is
simply the probability of event B.
Conditional Probability for 2 Independent Events
P(B|A) = P(B)
Example: - What is the probability that the second toss
of a fair coin will result in heads, given that heads
resulted on the first toss?
P(H2|H1) = 0.5
For 2 independent events, the result of the first toss
have absolutely no effect on the results of the second toss.

19. Probabilities under conditions of Statistical
Dependence
Statistical Dependence exists when the probability
of some event is dependent on or affected by the
occurrence of some other event.
The types of probabilities under statistical
dependence are:
• Marginal
• Joint
• Conditional

20. Example
Assume that a box contains 10 balls distributed as follows: -
3 are colored & dotted
1 is colored & striped
2 are gray & dotted
4 are gray & striped
Event Probability of Event
1 0.1
Colored & Dotted
2 0.1
3 0.1
4 0.1Colored & Striped
5 0.1
Gray & Dotted
6 0.1
7 0.1
Gray & Striped
8 0.1
9 0.1
10 0.1

21. Example: Marginal Probability - Statistically Dependent
It can be computed by summing up all the joint events
in which the simple event occurs.
Compute the marginal probability of the event colored.
It can be computed by summing up the probabilities of
the two joint events in which colored occurred:
P(C) = P(CD) + P(CS)
= 0.3 + 0.1
= 0.4

22. Example: Joint Probability - Statistically Dependent
Joint probabilities under conditions of statistical
dependence is given by
Joint probability for Statistically Dependent Events
P(BA) = P(B|A) * P(A)
•What is the probability that this ball is dotted and
colored?
Probability of colored & dotted balls =
P(DC) = P(D|C) * P(D)
= (0.3/0.4) * 0.5
= 0.375

23. Example: Conditional Probability - Statistically
Dependent
Given A & B to be the 2 events then,
Conditional probability for Statistically Dependent Events
P(BA)
P(B|A) = ----------
P(A)
Probability of event B given that event has occurred
P(B|A)

24. What is the probability that this
ball is dotted, given that it is
colored?
The probability of drawing any
one of the ball from this box is
0.1 (1/10) [Total no. of balls in
the box = 10].

25. We know that there are 4 colored balls, 3 of which
are dotted & one of it striped.
P(DC) 0.3
P(D|C) = --------- = ------
P(C) 0.4
= 0.75
P(DC) = Probability of colored & dotted balls
(3 out of 10 --- 3/10)
P(C) = 4 out of 10 --- 4/10

26. Type of
probability
Symbo
l
Formula
under
statistical
Independen
ce
Formula
under
statistical
dependence
1. MarginalP(A) P(A) P(A)
2. Joint P(AB)P(A) x P(B)P(A|B) x P(B)
3.
Conditional
P(A|B) P(A) P(AB)
P(B)

27. Revising Prior Estimates of Probabilities:
Bayes’ Theorem
A very important & useful application of conditional
probability is the computation of unknown
probabilities, based on past data or information.
When an event occurs through one of the various
mutually disjoint events, then the conditional
probability that this event has occurred due to a
particular reason or event is termed as Inverse
Probability or Posterior Probability.
Has wide ranging applications in Business & its
Management.

28. Since it is a concept of revision of probability based
on some additional information, it shows the
improvement towards certainty level of the event.
Example 1: - If a manager of a boutique finds that
most of the purple & white jackets that she thought
would sell so well are hanging on the rack, she must
revise her prior probabilities & order a different color
combination or have a sale.
Certain probabilities were altered after the people
got additional information. New probabilities are
known as revised, or Posterior probabilities.

29. Bayes Theorem
If an event A can occur only in conjunction with n
mutually exclusive & exhaustive events B1, B2, …, Bn, & if A
actually happens, then the probability that it was
preceded by an event Bi (for a conditional probabilities
of A given B1, A given B2 … A given Bn are known) & if
marginal probabilities P(Bi) are also known, then the
posterior probability of event Bi given that event A has
occurred is given by:

P(A | Bi). P(Bi)
P(Bi | A) = ----------------------
∑ P(A | Bi). P(Bi)

30. Remarks: -
The probabilities P(B1), P(B2), … , P(Bn) are termed
as the ‘a priori probabilities’ because they exist
before we gain any information from the
experiment itself.
The probabilities P(A | Bi), i=1,2,…,n are called
‘Likelihoods’ because they indicate how likely the event
A under consideration is to occur, given each & every
a priori probability.
The probabilities P(Bi | A), i=1, 2, …,n are called
‘Posterior probabilities’ because they are determined
after the results of the experiment are known.

31. Bayes’ Formula

32. Problem
In a bolt factory machines A, B, & C manufacture
respectively 25%, 35%, & 40% of the total. Of their
output 5%, 4%, 2% are defective bolts. A bolt is
drawn at random from the product & Is found to be
defective.
What are the probabilities that it was manufactured
by
machines A, B & C?

33. Solution
Let E1, E2, E3 denote the events manufactured by
machines A, B & C respectively.
Let E denote the event of its being defective.
P(E1) = 0.25; P(E2) = 0.35; P(E3) = 0.40;
Probability of drawing a defective bolt
manufactured by machine A is P(E|E1) = 0.05
Similarly P(E|E2) = 0.04;P(E|E3) = 0.02
Probability that defective bolt selected at random is
manufactured by machine A is given by

34. P(E1). P(E|E1)
P(E1|E) = ------------------------
∑ P(E1). P(E|E1)

i=1 to 3
0.25*0.05
= ----------------------------------------------
0.25*0.05 + 0.35*0.04 + 0.40*0.02
= 25/69
Similarly P(E2|E) = 28/69
= [(0.35*0.04)/(.25*.05+.35*.04+.40*.02)]
P(E3|E) = 16/69 =
[(0.40*0.02)/(.25*.05+.35*.04+.40*.02)]

35. Suppose that one person in 100, 000 has a
particular rare disease for which there is a fairly
accurate diagnostic test. This test is correct 99%
of the time when to someone with the disease;
it is correct 99.5% of the time when given to
someone who does not have the disease.
Given this information can we find
(a) the probability that someone who tests
positive for the disease has the disease?
(b) the probability that someone who tests
negative for the disease does not have the
disease?
Should someone who tests positive be very
concerned that he or she has the disease?

36. Glossary of terms
Classical Probability: It is based on the idea that
certain occurrences are equally likely.
Example: - Numbers 1, 2, 3, 4, 5, & 6 on a fair die are
each equally likely to occur.
Conditional Probability: The probability that an event
occurs given the outcome of some other event.
Independent Events: Events are independent if the
occurrence of one event does not affect the
occurrence of another event.
Joint Probability: Is the likelihood that 2 or more events
will happen at the same time.
Multiplication Formula: If there are m ways of doing
one thing and n ways of doing another thing, there are
m x n ways of doing both.

37. Mutually exclusive events: A property of a set of categories
such that an individual, object, or measurement is included in
only one category.
Objective Probability: It is based on symmetry of games of
chance or similar situations.
Outcome: Observation or measurement of an experiment.
Posterior Probability: A revised probability based on additional
information.
Prior Probability: The initial probability based on the present
level of information.
Probability: A value between 0 and 1, inclusive, describing the
relative possibility (chance or likelihood) an event will occur.
Subjective Probability: Synonym for personal probability.
Involves personal judgment, information, intuition, & other
subjective evaluation criteria.
Example: - A physician assessing the probability of a
patient’s recovery is making a personal judgment based
on what they know and feel about the situation.

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