Probability Foundation Course - Topics Covered

Introduction to Prob:
- The essential understanding of Permutations, Combinations and Percentiles
Multiple Event Probability:
- The Addition Rule of Probability
- The Multiplication Rule of Probability
- Conditional Probability
- Probability Trees
- Bayes Theorem
Discrete and Continious Probabilities

1. Quick Introduction to Probability:

Probability - Odds of a particular event happening over all possible outcomes
Prob = ^{# of Desired Outcomes} / _{Total # of Possible Outcomes}
Basic types of Probability:
- Classical Probability
- Empirical Probability
- Subjective Probability
Both Classical and Empirical Probabilities are Objective Probabilities where
- in Classical: The odds won’t change. They are based on formal reasoning. They based on established events/theory. E.g.: flipping a fair coin, picking a card from a usual pack of 52 cards
- in Emprical: The odds are based on experimental or historical data E.g.: What is the chance of a particular player scoring above 50 runs this match? (this can be determined by the historical data of that player)
Subjective probabilities are based on personal beliefs
Types of odds:
- Even Odds | Equally likely events
  - E.g.: Flipping a coin or Rolling a fair die
- Weighted Odds | Events with unequal chances of occuring
  - E.g.: Chances of occurrence of rain in Chennai today

2. How to count number of possible outcomes?

Permutations: If interested in the order of things
- E.g.: What is the chance for Students A and B to win top 2 prizes of a competition from a class of 10; There are 8 contestants and 3 prizes. How many different possible outcomes are there?
- nPk = ^n! / _(n-k)!
Combinations: If order is not important
- E.g.: What are the number of ways you can pick 4 members from a team of 12 members?
- nCk = ^n! / _k!*(n-k)!
Q1) There are 8 contestants and 3 prizes. How many different possible outcomes are there?
- “Permutations” problem because order is important
- 8! / (8-3)! = 8 X 7 X 6 = 336
Q2A) In how many ways can you “pick” 4 member team from a total 12 members?
- “Combinations” problem
- 12! / 4! * (12-4)! = 495
Q2B) In how many of these 495 combinations, do sisters Layla and Olivia join the same team?
- Assuming the sisters have already been inducted into the team,there are -10! / 2! * (10-2)!=45 ways to pick the remaining two members

Percentile Rank = Percentile rank of a given score is the percentage of scores in its freq distribution that are less than that score

  PR =( CF' + (0.5 X F) ) / N * 100
      CF' = Cumulative Freq (excluding the current score) = Count of all scores less than the score of interest
      F  = Freq of the score of interest 
      N = Total number of scores in the distribution

3. Multiple Event Probabilities

E.g. of single event probabilities:
Heads or Tails:
- Heads or Tails
- Rain or No Rain

E.g. of Multiple event probabilities:
Sports Example:
- 30% chance of a player scoring a goal
- 40% chance of that player’s team winning

Q) Is there a relationship between the team winning and this player’s success in scoring?

Healthcare example:
- 1/ 10K people gets a particular rare disease
- Test is accurate only 98% of the time

Q) What are the chances a person who is tested positive is actually false positive?

Employment example:
- Only 4 out of 20 get an interview
- What is the prob that friends Mohan and Lily both get a slot in the interview?
- What if Mohan gets one of the 4 slots, what is the prob of Lily securing one of the other 3 slots?

More Probability Tools:
- Conditional Probability
- Dependent vs Independent events
- Probability Trees and Bayes Theorem (both useful in managing multiple event scenarios)

4. More Probability Questions (using above concepts):

Q1. there are 6 people getting rewarded, what are the chances that 2 people - X and Y - win the gift?

Total number of outcomes = 
6! / (4! * 2!) = 15 

Number of outcomes where X, Y are both winning = [(X,Y), (Y,X)]
Order does not matter, hence number of outcomes of X and Y winning = 1
= 1/15 = 0.0667

Q2.What is the probability of rolling two dice with each die throwing 1?

1/6 * 1/6 = 0.0278 = 0.0278 2.7%

Q3. There are 10 cards with 3 of them having X on them. What are the odds that 2 cards picked at random have X on them?

Solution method 1: `Combinations` approach
Total number of combinations = 10C2 
= 10! /(8! x 2!) = 45 

Total number of combinations with 2 X on them = 3C2 = 3! / (2! * 1!) = 3 

Prob of picking 2 cards at random  where both are X = 3/45 = 0.0667 


Solution method 2: `Conditional Prob` approach

Chances of picking X card in attempt 1 = 3/10 = 0.3 

Chances of picking an X card in attempt 2 as well = 2/9 = 0.2222 

Chance of picking 2 X cards (2 dependent events) = 0.3 * 0.2222 = 0.0667

Q4. What are the chances that a medical test taken was false positive? (it is a conditional probability where given that the result is positive, what are the chances that it is false)

Disease or No Disease
Positive or Negative Test Result (could be false positive or true negative also)

Stats:
    1. only 1 in 10,000 has disease
    2. those with disease test positive 99% of time (that remaining 1% is False Negative or Type II)
    3. 2% of healthy paitents will test positive


Tree 
    - Stage 1:
    Disease 1/10000 = 0.0001 
        - Stage 2:
        Positive: 99/100 = 0.99 
        Negative: 1/100 = 0.01 

    No Disease 9999/10000 = 0.9999 
        Positive: 2/100 = 0.02 
        Negative: 98/100 = 0.98 

    - Total share of people tested positive = (0.0001 * 0.99 +   0.9999 * 0.02) = 0.0201 
    - Total share of people tested positive false 
    = 0.9999 * 0.02 = 0.02 
    - Total share of people tested true positive 
= 0.99 * 0.0001


    Prob of false positive = Share of False Positive / Total # of Positives = 0.02 / 0.0201 = 0.995 = 99.5%

    Prob of True Positive = Share of True Positive / Total # of positives = (0.99 * 0.0001) / 0.0201  = 0.0049 = 0.5%

Q5. 70% of the population has brown eyes, 30% do not have brown eyes. 60% of the population requires reading glasses, 40% do not need reading glasses. In a city of 10,000 people, how many would both not have brown eyes and not require reading glasses?

Independent events - just multiply the prob. 
P(not_brown_eyes) * P(not_require_glasses) = 0.3 * 0.4 = 0.12 

0.12 * 10000 = 1200

Q6. There are two stacks of cards. Each stack has 4 cards. Each stack has a card with the numbers 1, 2, 3, and 4. There are 16 possible outcomes. You will be allowed to take one card from each stack. Two cards total. What is the probability of drawing at least one card with a 4 from either deck

Total_num_of_cards = 8
Total_num_of_possible_outcomes = 4C1 * 4C1 = 16

= 1/4 * 1 + 1 * 1/4 - 1/16 = 0.4375

Q7. Suppose you have 3 coins, each with heads on one side and tails on the other. There are 8 possible outcomes. What is the probability that when all three coins are flipped at least 2 coins will result in heads?

total_num_of_outcomes = 2 * 2 * 2 = 8 


HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

Q8. There are ten people in a class. Ari and Jamaal are twins in this class. At random two people will be chosen as the class representatives. What are the odds that Ari and Jamaal will both be chosen?

Total_num_of_combinations = 10C2 = 10! / (8! * 2!) = 45 

Only one combination is Ari and Jammal = 1/45 = 0.0222

Q9. A company has 1000 employees. 70% get the flu vaccine. 95% of those that get the vaccine do NOT get the flu, 5% get the flu. 30% do not get the flu vaccine. 80% of those that do not get the vaccine do not get the flu; 20% that do not get the vaccine do get the flu. How many of the 1000 employees get the vaccine but still get the flu?

700 Vaccinated
-- NoFlu: 0.95 * 700 = 665 
--Flu: 0.05 * 700 = 35 
300 Non vaccinated
-- NoFlu: 0.8*300 = 240 
-- Flu: 0.2 * 300 = 60

Source: - LinkedIn Courses - Statistics Foundations: Probability and Statistics Foundations: The Basics | refer