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probability equations

Module by: Adan Galvan. E-mail the author

Summary: A collection of probability equations.

Probability equations

Simple Probability

Prx=Favorable OutcomesPossible Outcomes x Favorable Outcomes Possible Outcomes
(1)

Exercise 1

What is the probability that a card drawn at random from a deck of cards will be an ace?

Solution

Since of the 52 cards in the deck, 4 are aces, the probability is 4/52.

Conditional Probability

A conditional probability is the probability of an event given that another event has occurred. For example, what is the probability that the total of two dice will be greater than 8 given that the first die is a 6? This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown in the section on simple probability. There are 6 outcomes for which the first die is a 6, and of these, there are four that total more than 8 (6,3; 6,4; 6,5; 6,6). The probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3. More formally, this probability can be written as: Prtotal>8| Die 1=6 =2/3 Die 1 6 total 8 23 . In this equation, the expression to the left of the vertical bar represents the event and the expression to the right of the vertical bar represents the condition. Thus it would be read as "The probability that the total is greater than 8 given that Die 1 is 6 is 2/3." In more abstract form, PrA| B B A is the probability of event A given that event B occurred.

Probability of A and B

Probability of A and B: Independent Variables

PrAB=PrAPrB A B A B
(2)

Exercise 2

What is the probability that a fair coin will come up with heads twice in a row?

Solution

Two events must occur: a head on the first toss and a head on the second toss. Since the probability of each event is 1/2, the probability of both events is: 1/2×1/2=1/4 12 12 14 .

Probability of A and B: Dependent Variables

PrAB=PrAPrB| A A B A A B
(3)

Exercise 3

If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?

Solution

Event A is that the first card is an ace. Since 4 of the 52 cards are aces, PrA=4/52=1/13 A 452 113 . Given that the first card is an ace, what is the probability that the second card will be an ace as well? Of the 51 remaining cards, 3 are aces. Therefore, PrB| A =3/51=1/17 A B 351 117 and the probability of A and B is: 1/13×1/17=1/221 113 117 1221 .

Probability of A or B

Probability of A or B: Mutually Exclusive Variables

PrAB=PrA+PrB A B A B
(4)

Probability of A or B:Not Mutually Exclusive Variables

PrAB=PrA+PrBPrAB A B A B A B
(5)

Binomial Distribution

The binomial probability for obtaining rr successes in NN trials is:

Prr=N!r!(Nr)!πr1πNr r N r N r π r 1 π N r
(6)
where Prr r is the probability of exactly rr successes, NN is the number of events, and pp is the probability of success on any one trial. This formula assumes that the events:
  1. are dichotomous (fall into only two categories)
  2. are mutually exclusive
  3. are independent and
  4. are randomly selected

Exercise 4

Consider this simple application of the binomial distribution: What is the probability of obtaining exactly 3 heads if a fair coin is flipped 6 times?

Solution

For this problem, N=6 N 6 , r=3 r 3 , and π=0.5 π 0.5 . Therefore, Pr3=6!3!(63)!0.5310.563=6×5×4×3×2(3×2)(3×2)(0.125×0.125)=0.3125 3 6 3 6 3 0.5 3 1 0.5 6 3 6 5 4 3 2 3 2 3 2 0.125 0.125 0.3125

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