Rules of Probability
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When considering a given procedure, events can be and often are related to one another. If we know the probabilities of one or more events, we can find the probabilities of other events by relying on the ones we already know.
The simplest example of this is the complement rule.
The Complement of an Event:
The complement of Event A is denoted A’ (read as “A prime”) and consists of everything in the Sample Space except Event A.
Since the two together comprise the sample space, their probabilities sum to 1. (Remember that in probabilities, 1 = certainty. If all possible outcomes are included, it is certain that the eventual outcome will be one of those listed.)
We express this rule in this notation: P(A) + P(A’) = 1
The probability of the complement is, then, found by subtracting the known probability of Event A from 1. This is expressed in this notation: P(A’) = 1 – P(A) For ex. The Wall Street Journal reports that .33 of all small businesses fail within the first 2 years. This leaves the probability for small businesses lasting 2 years or more at .67. Because 1 - .33 = .67. (It can also help to think of A’ as “not A”)
The Union of Two Events:
The union of two events consists of all outcomes in the same space (S) that are contained in either Event A or Event B or both.
We express the union of the two sets in this notation: P(A ᴜ B). This is often read as A or B since it means that either or both events occur. Ex. Choosing a card from a random deck of playing cards: If Q is the event that we will choose a queen and R is the event that we will draw a red card, Q ᴜ R consists of getting either a queen or a red card.
The Intersection of Two Events: The intersection of two events is the event consisting of all outcomes in the sample space S that are contained in both Event A and Event B.
We express the intersection of events in this notation: P(A ∩ B). This is often read as A and B since it means that both events occur. This is also called the joint probability. Ex. If Q is the event that we draw a queen and R is the event that we draw a red card, Q ∩ R is the event that we will draw a card that is both red and a queen. Therefore, out of 52 cards, Q ∩ R consists of 2 cards, the Queen of Hearts and Queen of Diamonds.
General Law of Addition: The probability of the union of two events A and B, is the sum of their probabilities less the probability of their intersection:
Expressed as: P(A á´œ B) = P(A) + P(B) – P(A ∩ B)
If we just added the probabilities of the union of A & B, we would count their intersection twice, so we must subtract the intersection to avoid overstating the probability of A ᴜ B. In the card example:
Queen: P(Q) = 4/52 (there are four queens per deck)
Red: P(R) = 26/52 (there are 26 red cards per deck)
Queen & Red P(Q ∩ R ) = 2/52 (there are 2 red queens per deck)
Therefore,
Queen or Red: P(Q á´œ R) = P(Q) + P(R) – P (Q ∩ R )
= 4/52 + 26/52 - 2/52
= 28/52 = .5385 or a 53.85% chance of drawing a red card OR a queen.
Mutually Exclusive Events:
Events A and B are mutually exclusive (or disjoint) if their intersection is the empty set (a set that contains no elements). In other words, one event precludes the other from occurring. The null set is expressed with the notation: φ.
If (A ∩ B) = φ¸then P(A ∩ B)= 0
Ex. A is the event that an Applebee’s customer will finish her lunch in fewer than 30 minutes. B is the event that she will finish her lunch in 30 minutes or more. The two events have no overlap. Mutually exclusive events may not cover all possibilities. Their distinguishing factor is that the two sets of possibilities that are drawn have no overlap between them.
Special Law of Addition:
If A and B are mutually exclusive events, then P(A ∩ B) = 0 and the addition law reduces to P(A á´œ B) = P(A) + P(B). There is no overlap of events in mutually exclusive events. There is only an “either/or” situation. So, it makes sense that we no longer subtract the overlap. Ex. If we look at a person’s age, and P(under 21) = .28 and P(over 65) = .12, so P(under 21 or over 65) = .28 + .12 = .40, since these two sets do not overlap.
Collectively Exhaustive Events:
Events are collectively exhaustive if their union is the entire sample space S (i.e., all the events that can possibly occur). Two mutually exclusive, collectively exhaustive events are binary (ordichotomous) events. Ex. A car repair is either covered by the warranty (A) or not covered by the warranty (A’).
There can be mutually exclusive, collectively exhaustive events can be numbered more than two. Ex. Walmart customers can pay by check (A), credit card (B), cash (C) or money order (D).
Categorical data can be made into binary data by defining the second category as everything not in the first category. Ex. A student either is a senior (A) or he is not (B). So, all juniors, sophomores and freshmen are in B.
Conditional Probability
The probability of event A given that event B has occurred is a conditional probability. The notation used to express this is: P(A|B). The vertical line is read as “given”.
Conditional probability is found by dividing the probability of the intersection by the probability of the known set. In other words:
P(A|B) = P(A ∩ B)/P(B) (for P(B) > 0) When B is the known probability.
The intersection (A ∩ B) is the part of B that is also in A. The ration of the relative size of the set A ∩ B is the conditional probability.
High School Drop Out Example
Of the population aged 16-21 and not in college, 13.50 percent are unemployed, 29.05 percent are high school dropouts and 5.32 percent are unemployed high school dropouts. What is the conditional probability that a member of this population is unemployed, given that the person is a high school dropout? To answer this question, define:
U = the event that the person is unemployed
D = the event that the person is a high school dropout
This “story problem” contained three facts: P(U) = .1350; P(D) = .2905; P(U ∩ D) = .0532
So by the formula above: P(A|B) = P(A ∩ B)/P(B), the conditional probability of an unemployed youth given that the person dropped out of high school is:
P(U|D) = P(U∩D)/P(D) = .0532/.2905 = .1831 or 18.31%
The conditional probability of being unemployed is P(U|D) = .1831 (18.31 percent) which is greater than the unconditional probability of being unemployed P(U) = .1350(13.5 percent). In other words, knowing that someone is a high school dropout alters the probability that the person is unemployed.
Odds of an Event
Statisticians usually speak of probabilities rather than odds, but in sports and games of chance, we often hear odds quoted. We define the odds in favor of an event (ex. A) as the ratio of the probability that the event will occur to the probability that it will not occur. In other words will occur/will not occur, also known as P(A)/P(A’). And those are defined as the Probability of A [P(A)] divided by the probability of “A prime” [P(A’)] or “not A”. We calculate that in this notation:
P(A)/1-P(A)
Odds against an event: Reverse the ratio to P(A’)/P(A) (a/k/a “Not A” divided by A)
For a pair of dice, the probability of rolling a seven is 6/36 or 1/6, so the odds in favor of rolling a seven would be:
Odds = P(rolling seven)/1-P(rolling seven) = 1/6 / 1 - 1/6 = 1/6 / 5/6 = 1/5
This means that, on average, for every time we roll a seven, there will be five times that we do not roll a seven. The odds are 1 to 5 in favor of rolling a seven or, conversely stated, 5 to 1 againstrolling a seven.
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