Probability

strategy- intersection

look for the intersections and see if using the multiplicative property to rewrite them as P( A | B ) P(B) will make them simpler to deal with

How important theorems interact with conditional probability

Important Theorems 1. Law of Total Probability: Enables us to calculate unconditional probabilities by using conditional probabilities. 2. Bayes' Theorem: Enables us to calculate conditional probabilities if we have information on the reverse conditional probabilities.

mutual exclusivity vs independence

Mutually exclusive events cannot happen at the same time. For example: when tossing a coin, the result can either be heads or tails but cannot be both. Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).Sep 22, 2014

synonym for mutually exclusive

disjoint

when are disjoint events independent

disjoint events are independent only if P(A) = 0 and P(B) = 0.

difference between independence and disjointness

if disjoint, P(A intersect B) = 0

Two events are mutually exclusive

if they can't both happen

synonym for disjoint

mutually exclusive

Mutual Independence

Definition A set of events {A1, A2, . . . , An} are mutually independent if for any subset {Ai, Aj, . . . , Am}, we have P(Ai ∩ Aj ∩ · · · ∩ Am) = P(Ai)P(Aj)· · · P(Am). Equivalently for {A¯i, A¯j, . . . , A¯m}, where A¯i is Ai or A˜i ,P(A¯i ∩ A¯j ∩ · · · ∩ A¯m) = P(A¯i)P(A¯j)· · · P(A¯m).

the multiplicative rule of conditional probability:

From this definition, we get the multiplicative rule of conditional probability: P(A ∩ B) = P(A|B)P(B) = P(B|A)P(A)

the Axioms of probability:

1. P(A) ≥ 0, for any event A 2. P(S) = 1 3. If A and B are mutually exclusive (disjoint) events, then P(A ∪ B) = P(A) + P(B)

Example A nationwide poll determines that 72% of the American population loves eating pizza. If two people are randomly selected from the population, what is the probability that the first person loves eating pizza, while the second one does not?

Solution. Let A be the event that the first person loves pizza, and let B be the event that the second person loves pizza. Because the two people are selected randomly from the population, it is reasonable to assume that A and B are independent events. If A and B are independent events, then A and B' are also independent events. Therefore, P(A ∩ B'), the probability that the first person loves eating pizza, while the second one does not can be calculated by multiplying their individual probabilities together. That is, the probability that the first person loves eating pizza, while the second one does not is: P(A ∩ B') = 0.72 × (1 − 0.72) = 0.202

Independence

The situation where events provide no information about each other

Why can an event not be both mutually exclusive and independent (except in the one case in which this can occur)

because if ME and A did occur this gives us info about B, that it wont occur as the events cannot occur together, thus A is giving us info on B and the events are not independent.

mutual independence implies.. BUT

pairwise independence but you still have to check for pairwise independence

Pairwise Independence

Definition A set of events {A1, A2, . . . , An} are said to be pairwise independent if for every pair {Ai , Aj}, i 6= j, we have P(Ai ∩ Aj) = P(Ai)P(Aj).

formal definition of independent events

Definition. Events A and B are independent events if the occurrence of one of them does not affect the probability of the occurrence of the other. That is, two events are independent if either: P(B|A) = P(B) (provided that P(A) > 0) or: P(A|B) = P(A) (provided that P(B) > 0).

formal definition of mutual independence

Definition. Three events A, B, and C are mutually independent if and only if the following two conditions hold: (1) The events are pairwise independent. That is, P(A ∩ B) = P(A) × P(B) and... P(A ∩ C) = P(A) × P(C) and... P(B ∩ C) = P(B) × P(C) (2) P(A ∩ B ∩ C) = P(A) × P(B) × P(C)

Alternate definition of independence

Now, since independence tells us that P(B|A) = P(B), we can substitute P(B) in for P(B|A) in the formula given to us by the multiplication rule: P(A ∩ B) = P(A) × P(B |A) yielding: P(A ∩ B) = P(A) × P(B). This substitution leads us to an alternative definition of independence. Definition. Events A and B are independent events if and only if : P(A ∩ B) = P(A) × P(B) Otherwise, A and B are called dependent events. Recall that the "if and only if" (often written as "iff") in that definition means that the if-then statement works in both directions. That is, the definition tells us two things: If events A and B are independent, then P(A ∩ B) = P(A) × P(B). If P(A ∩ B) = P(A) × P(B), then events A and B are independent.

If A and B are independent P(A I B ) equals

P( A)

Events A and B are independent if:

P(A intersect B) = P(A) * P(B)

the Inclusion-Exclusion Principle:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

Conditional Probability Definition

P(A|B) = P(A ∩ B) / P(B) provided P(B) > 0.

A recent survey of students suggested that 10% of Penn State students commute by bike, while 40% of them have a significant other. Based on this survey, what percentage of Penn State students commute by bike and have a significant other?

Solution. Let's let B be the event that a randomly selected Penn State student commutes by bike, and S be the event that a randomly selected Penn State student has a significant other. If B and S are independent events (okay??), then the definition tells us that: P(B ∩ S) = P(B) × P(S) = 0.10 × 0.40 = 0.04 That is, 4% of Penn State students commute by bike and have a significant other. (Is this result at all meaningful??)

Gambler's fallacy

The moral of the story is to be careful not to fall prey to "Gambler's Fallacy," which occurs when a person mistakenly assumes that a departure from what occurs on average in the long term will be corrected in the short term. In this case, a person would be mistaken to think that just because the coin was departing from the average (half the tosses being heads and half the tosses being tails) by getting eight straight tails in row that a head was due to be tossed. A classic example of Gambler's Fallacy occurred on August 18, 1913 at the casino in Monte Carlo, in which: ... black came up a record twenty-six times in succession [in roulette]. ... [There] was a near-panicky rush to bet on red, beginning about the time black had come up a phenomenal fifteen times. In application of the maturity [of the chances] doctrine, players doubled and tripled their stakes, this doctrine leading them to believe after black came up the twentieth time that there was not a chance in a million of another repeat. In the end the unusual run enriched the Casino by some millions of francs. [Source: Darrell Huff & Irving Geis, How to Take a Chance (1959), pp. 28-29.] A joke told among statisticians demonstrates the nature of the fallacy. A man attempts to board an airplane with a bomb. When questioned by security, he explains that he was just trying to protect himself: "The chances of an airplane having a bomb on it are very small," he reasons, "and certainly the chances of having two are almost none!" A similar example is in the book by John Irving called The World According to Garp, in which the hero Garp decides to buy a house immediately after a small plane crashes into it, reasoning that the chances of another plane hitting the house have just dropped to zero.

Consider a roulette wheel that has 36 numbers colored red (R) or black (B) according to the following pattern: data and define the following three events: Let A be the event that a spin of the wheel yields a RED number = {1, 2, 3, 4, 5, 10, 11, 12, 13, 24, 25, 26, 27, 32, 33, 34, 35, 36}. Let B be the event that a spin of the wheel yields an EVEN number = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36}. Let C be the event that a spin of the wheel yields a number no greater than 18 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}. Now consider the following two questions: Are the events A, B, and C "pairwise independent?" That is, is event A independent of event B; event A independent of event C; and B independent of event C? Does P(A ∩ B ∩ C) = P(A) × P(B) × P(C)?

not mutually independent

One ball is drawn randomly from a bowl containing four balls numbered 1, 2, 3, and 4. Define the following three events: Let A be the event that a 1 or 2 is drawn. That is, A = {1, 2}. Let B be the event that a 1 or 3 is drawn. That is, B = {1, 3}. Let C be the event that a 1 or 4 is drawn. That is, C = {1, 4}. Are events A, B, and C pairwise independent? Are they mutually independent?

pairwise independent! Intersection between each set contains "1" thus intersection between each pair contains one thing out of four total things in the sample space. thus 1/4 and 1/4 = 0.5 * 0.5 with 0.5 being the prob of a b or c occuring individually thus pairwise. Not mutual because 1/4 is AintCintB but (1/2)^3 is not equal to (1/4).

The red even odd problem. Why it is independent

there are just as many even as odd numbers and just as many even red numbers as there are odd red numbers, thus knowing that it is even, does not increase the chances of it being a red number, and knowing that it is red does not increase the chances of it being even.

strategy # 2

think of how to rewrite word problems using "and" or "or"