- Probability refers to chance or likelihood of a particular Event taking place
- Event is an outcome of the Experiment
- Experiment is a process that is performed to understand or observe possible Outcomes
- Set of all Outcomes of the Experiment is called the Sample Space
PROBLEM STATEMENT
Consider the following, in a manufacturing unit three parts from the assembly are selected at random. You are observing whether they are defective or not. Determine
- Sample Space
- The Event of getting at least two defective parts
If D = Defective and G = Non-Defective, then
- Sample Space S = {DDD, DDG, DGD, DGG, GDD, GDG, GGD, GGG}
- E = {DDD, DDG, DGD, GDD}
PROBABILITY
- Probability of an Event A is defined as a ratio of two numbers m and n
- P(A) = m/n, where m = number of outcomes favoring event A, n = total number of outcomes
- P(A) is always between the range >= 0 and <= 1
- Two Events A and B are said to be Mutually Exclusive if at any given point only one Event outcome is possible.
- For example, if one card is picked at random from the well shuffled pack of cards, the event of King or Queen being picked is Mutually Exclusive. The selected card is either a King or a Queen, it cannot be both.
- This can be thought of as a OR operation
- Two Events A and B are said to be Independent if at any given point occurrence of A is in no way influenced by the occurrence of B and vice versa.
- For example, if one coin is tossed multiple times, occurrence of Head in one toss is in no way influenced/dependent on earlier coin toss.
- This can be thought of as a AND operation.
ADDITION RULE - MUTUALLY EXCLUSIVE EVENTS
- The rule states that the probability of the union of A and B is determined by adding the probability of events A and B
- Represented as shown below
- For example, from a pack of well shuffled cards, a card is picked at random. What is the probability that the selected card is a King or a Queen?
- Total Events = 4 Kings + 4 Queens
- Total Sample Space = 52 cards
- Probability of King being picked = P(King) = 4/52 = 1/13
- Probability of Queen being picked = P(Queen) = 4/52 = 1/13
- Probability of King or Queen picked = P(King) + P(Queen) = 2/13
ADDITION RULE - NOT MUTUALLY EXCLUSIVE EVENTS
- The rule states that the probability of the union of A and B is determined by adding the probability of events A and B and then subtracting the probability of the intersection of A and B
- Represented as shown below
- For example, from a pack of well shuffled cards, a card is picked at random. What is the probability that the selected card is a King or a Diamond?
- Total Events = 3 Kings + 3 Diamonds + 1 King and Diamond
- Total Sample Space = 52 cards
- Probability of King being picked = P(King) = 4/52
- Probability of Diamond being picked = P(Diamond) = 13/52
- Probability of King and Diamond being picked = P(KingDiamond) = 1/52
- Probability of King or Diamond picked = P(King) + P(Diamond) - P(KingDiamond) = (4 + 13 - 1)/52 = 16/52 = 4/13
MULTIPLICATION RULE - INDEPENDENT EVENTS
- The rule states that the probability of the intersection of A and B is determined by product of the probability of events A and B
- This rule can be extended to two or more events
- Represented as shown below
- For example, suppose the probability of you getting a A grade in Statistics is given as 0.6 and the probability of you getting a A grade in Business Finance is given as 0.5. Assuming these courses are independent, compute the probability of getting A in both the courses.
- Probability of A grade in Statistics is given as = P(S) = 0.6
- Probability of A grade in Business Finance is given as = P(F) = 0.5
- Probability of A grade in both Statistics and Business Finance = P(S) * P(F) = 0.6 * 0.5 = .30
MULTIPLICATION RULE - NOT INDEPENDENT EVENTS
- The rule states that the probability of the intersection of A and B is determined by product of either
- The probability of event A and probability of event B provided A has already occurred
- The probability of event B and probability of event A provided B has already occurred
- Represented as shown below
- For example, from the pack of cards, 2 cards are drawn in succession one after the other. After every draw, the selected card is not replaced. What is the probability that in both the draws you will get Spades?
- Probability of Spade in first draw = P(S1) = 13/52
- Probability of Spade in second draw provided Spade is already drawn and card not replaced = P(S2) = 12/51
- Probability of getting Spade in both draws = P(S1) * P(S2) = 13/52 *12/51 = 1/17
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