This document discusses probability concepts and their applications. It covers basic terminology like events, experiments, and types of probability such as classical, relative frequency, and subjective approaches. It also discusses concepts like marginal, joint, and conditional probabilities for both independent and dependent events. Several examples of probability problems and their solutions are provided to illustrate calculating probabilities of events.

1. Probability Concepts
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2. Applications
• Insurance Companies
• Medical Research
• Risk Calculation
• Product Management
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3. Basic Terminology in Probability
• Event
• Possible outcomes of doing an activity
• Experiment
• The activity that produces an event
• Mutuality of the events
• Exclusive: Independent events
• Inclusive: Dependent events
• Probability of occurance
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4. Check!!!!
• Collectively exhaustive list:
• Coin: (Head, Head); (Head, Tail); (Tail, Head); (Tail, Tail)
• Dice: (1,1); (1,2);…… (6,6)
• Probability of Occurrence in Dice:
• Rolling of x = P(x) =
= (Count of event occurances) / All Possible combinations
• Mutually Exclusive: From the Deck of cards
• Heart and Queen
• Ace and Even number
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5. Types of Probability
• Classical Approach
• Relative Frequency Approach
• Subjective Approach
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6. Classical Probability
• Probability:
Number of outcomes where the event occurs
Total Number of Possible outcomes
• Also a priori probability
Shortcomings:
• Small use cases
• No consideration of real time scenarios
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7. Relative Frequency of occurrence
Probability based on the past occurrences
• Death of an insured person
• Defaulting a loan at a finance company
Probability:
Occurences based on the earlier data
With the total occurrences
• More Data – More Accurate
Shortcomings:
• Insufficient Data
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8. Subjective Probabilities
• Subjective assessments, either from the previous occurrences or from
the extracted knowledge
• Classic Eg: Board Exam Results
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9. Probability Rules
P(A) = The probability of event A happening
P(A and B) = The probability of event A and event B happening
P (A or B) = The probability of event A or event B happening
Exclusive Events A and B = P(A or B) = P(A) + P(B)
Inclusive Events A and B = P(A or B) = P(A) + P(B) – P(A and B)
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10. Samples
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A
11
B
7
C
9
A B
C
8 5
5
4
2 9
6

11. Probabilities under conditions of statistical
independence
• Marginal
• Joint
• Conditional
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12. Marginal Probabilities of Independent Events
Simple Probability of distribution of all possible outcomes
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13. Joint Probabilities of Independent Events
• Marginal probability of A and B = P(AB)
• Probability of events A and B occurring together or in succession
• P(AB) = P(A) * P(B) ------- Proved
• 2 Coins tossed
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14. Conditional Probabilities of Independent
Events
Probability of occurrence of 2nd event (B) under the condition of
occurrence of 1st event (A) = P(B|A)
P(B|A) = P(B)
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15. Probabilities under conditions of statistical
dependence
• Marginal
• Conditional
• Joint
Bag of balls contains:
3 Coloured and Dotted
1 Coloured and Striped
2 Grey and Dotted
4 Grey and Striped
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16. Marginal Probabilities of dependent Events
• P(C) = P(CS)+P(CD)
• P(G) = P(GS)+P(GD)
• P(D) = P(CD)+P(GD)
• P(S) = P(CS)+P(GS)
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17. Conditional Probabilities of dependent Events
• Consider only coloured balls from
the bag
• Probability of dotted
= P(D|C) = ¾
• Probability of striped
=P(S|C) = ¼
• Consider all balls from the bag
• Probability of dotted and coloured
3 out of 10 = 3/10 = 0.3
• Probability of coloured balls
4 out of 10 = 4/10 = 0.4
• Probability of dotted
= P(DC)/P(C) = 0.3/0.4 = 3/4
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Hence, P(D|C) = P(DC)/P(C)

18. Joint Probabilities of dependent Events
P(D|C) =
P(DC)
P(C)
P(DC) = P(D|C) * P(C)
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19. Bayes Theorem
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P(B|A) =
P(AB)
P(A)
P(A|B) = P(B|A)*P(B)/P(A)

20. Problem 1:
A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from
the pack, not replaced and then another pen is drawn. What is the probability of
drawing 2 blue pens and 1 black pen?
Solution:
Probability of drawing 1 blue pen = 4/9
Probability of drawing another blue pen = 3/8
Probability of drawing 1 black pen = 3/7
Probability of drawing 2 blue pens and 1 black pen = 4/9 * 3/8 * 3/7 =
1/14
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21. Problem 2:
What is the probability of drawing a king and a queen consecutively from a deck of
52 cards, without replacement.?
Solution:
• Probability of drawing a king = 4/52 = 1/13
• After drawing one card, the number of cards are 51.
• Probability of drawing a queen = 4/51.
• Now, the probability of drawing a king and queen consecutively is
1/13 * 4/51 = 4/663
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22. Problem 3:
What is the probability of the occurrence of a number that is odd or less than 5
when a fair die is rolled?
Solution:
• Let the event of the occurrence of a number that is odd be ‘A’ and the event of
the occurrence of a number that is less than 5 be ‘B’. We need to find P(A or B).
• P(A) = 3/6 (odd numbers = 1,3 and 5)
• P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)
• P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)
• Now, P(A or B) = P(A) + P(B) – P(A or B)
• = 3/6 + 4/6 – 2/6
• P(A or B) = 5/6.
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23. Problem 4:
In a class, 40% of the students study math and science. 60% of the students study
math. What is the probability of a student studying science given he/she is already
studying math?
Solution:
• P(M and S) = 0.40
• P(M) = 0.60
• P(S|M) = P(M and S)/P(S) = 0.40/0.60 = 2/3 = 0.67
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24. Problem 5:
When two dice are rolled, find the probability of getting a greater number on the
first die than the one on the second, given that the sum should equal 8.
Solution:
• Let the event of getting a greater number on the first die be G.
• There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4),
(5,3),(6,2)}.
• And there are two ways where the number on the first die is greater than the one on the
second given that the sum should equal 8, G = {(5,3), (6,2)}.
• Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36.
• Now, P(G|sum equals 8) = P(G and sum equals 8)/P(sum equals 8)
• = (2/36)/(5/36)
• = 2/5
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25. Problem 6:
In a particular pain clinic, 10% of patients are prescribed narcotic pain killers. Overall, five
percent of the clinic’s patients are addicted to narcotics (including pain killers and illegal
substances). Out of all the people prescribed pain pills, 8% are addicts. If a patient is an
addict, what is the probability that they will be prescribed pain pills?
Solution:
• Step 1: Figure out what your event “A” is from the question. That information is in the italicized
part of this particular question. The event that happens first (A) is being prescribed pain pills.
That’s given as 10%.
• Step 2: Figure out what your event “B” is from the question. That information is also in the
italicized part of this particular question. Event B is being an addict. That’s given as 5%.
• Step 3: Figure out what the probability of event B (Step 2) given event A (Step 1). In other words,
find what (B|A) is. We want to know “Given that people are prescribed pain pills, what’s the
probability they are an addict?” That is given in the question as 8%, or .8.
• Step 4: Insert your answers from Steps 1, 2 and 3 into the formula and solve.
P(A|B) = P(B|A) * P(A) / P(B) = (0.08 * 0.1)/0.05 = 0.16
• The probability of an addict being prescribed pain pills is 0.16 (16%).
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