Probability theory has developed into a major discipline that supports applications in biology, physics, psychology, and business, to mention only a few. Probability deals with situations exhibiting uncertainty. Probability is a ratio .
In dealing with probability problems, we deal with experiments and events.
DEFINITIONS EXPERIMENTis any action which results in outcomes which occur unpredictably. SAMPLE SPACE of an experiment is a set S of the outcomes. EVENT is any subset of a sample space. |
In the experiment of rolling a single six sided die, the sample space consists of 6 events: 1, 2, 3, 4, 5, and 6.
NOTATION P denotes a probability. A, B and C denote specific events. P(A) denotes the probability of event A occuring |
For example, event A might represent the event of winning the power ball, and P(A) would be the probability of that event occuring.
P(A) is computed as follows:
If an event A is certain to occur, then P(A) = n/n = 1. If the event will never occur, then P(A) = 0/n = 0. Therefore, all other probabilities are assigned a number between 0 and 1. A summary of these three statements is 0 P(A) 1.
The probability of rolling a single 6 sided die and getting one of the numbers 1, 2, 3, 4, 5 or 6 is P(1,2,3,4,5,6) = 1. The probability of rolling a single six sided die and getting a 7 is P(7) = 0. The probability of rolling a 2 is P(2) = 1/6.
EXAMPLE: A single 6 sided die is rolled. State the sample space for this experiment and find the probability of rolling an even number.
SOLUTION The sample space is {1, 2, 3, 4, 5, 6}. Of these 6 equal likely outcomes (2, 4, and 6) correspond to the event of "rolling an even number". P(even number) = 3/6 = 1/2.
Two events A and B that cannot occur at the same time are called mutually exclusive events. In the above problem, the events of "rolling a 5" and "rolling an even number" are mutually exclusive.
Probability of Two Mutually Exclusive Events If events A and B are mutually exclusive, then P(A or B) = P(A) + P(B) |
At this point, you can see that probability involves counting. This counting process can be accomplished by using several techniques such as; tree diagrams, charts, tables, or mathematics.
Example: There are 3 choices of ice cream (vanilla, chocolate, strawberry) and 3 different toppings (M&M's, hot fudge, nuts-pecans) to choose from. How many choices are there?
From either of the methods you can see there are a 9 different choices of ice cream and toppings. In a mathematical sense we could reason that there are 3 choices for ice cream and with each of these choices there are 3 choices of toppings. Thus, there should be 3 x 3 = 9 choices.
COUNTING PRICIPLE - MULTIPLICATION If there are m ways of selecting an item from set A and n ways of selecting an item from set B, then there are m x n ways of selecting an item from set A and set B. |
Example: What is the probability that a couple with 3 children will have exactly 2 boys.
The sample space for this problem could be determined in seveal ways. Although, we also want to know how often the event of exactly 2 boys occurs.
1st Birth |
2nd Birth |
3rd Birth |
boy |
boy |
boy |
boy |
boy |
girl |
boy |
girl |
boy |
boy |
girl |
girl |
girl |
girl |
boy |
girl |
boy |
girl |
girl |
boy |
boy |
girl |
girl |
girl |
P(exactly 2 boys) = 3/8 .
Using the counting principle we have 2 choices for the first birth, 2 choices for the second birth and 2 choices for the third birth. Thus, 2 x 2 x 2 = 8 outcomes in this sample space. The number of specified outcomes now needs to be determined. A boy could be born on the 1st & 2nd births, 1st & 3rd births, or 2nd & 3rd births. Thus 3 favorable outcomes.
To account for the double counting of events that are not mutally exclusive we need to subtract the number of outcomes when the events occur at the same time.
Probability of Two Non- Mutually Exclusive Events If events A and B are not mutually exclusive, then P(A or B) = P(A) + P(B) – P(A AND B) |
Example: A single 6-sided die is rolled. What is the probability that the number is divisible by 2 or 3?
Solution: The sample space is {1, 2, 3, 4, 5, 6}. The sets of events or favorable outcomes is {2,4,6} and {3,6} . Notice, 6 is a member of both sets of events since 6 is divisible by both 2 and 3. Thus P(divisible by 2 or 3) = P(divisible by 2) + P(divisible by 3) – P(divisible by 2 and 3) = 3/6 + 2/6 -1/6 = 2/3
EVENTS OCCURING TOGETHER
Events can occur simultaneously or successively. The outcome of one event may or may not affect the outcome of the other. For example, suppose that a box contains 2 red, 3 green and 4 yellow marbles. If two marbles are drawn from the box, what is the probability of getting a red marble on both draws? If a marble is drawn and the placed back into the box, the probability of a red marble on the second draw is the same as the probability on the first draw. The P(2 reds) = 2/9 x 2/9 = 4/81. Now, if the first marble is not placed back in the box, the probability of getting a red marble on the second draw is affected. The P(2 reds) = 2/9 x 1/8 = 2/72 .
Two events are independent if the outcome of one has no effect on the outcome of the other. If the outcome of one event does affect the other then the events are dependent.
NOTATION: P(B given that A has already happened) = P(B | A) |
The probability that one event occurs given that another has occured is called conditional probability. P(B |A) denotes conditional probability.
DEPENDENT AND INDEPENDENT PROBABILITIES P(A and B) = P(A) • P(B) if A and B are independent P(A and B) = P(A) • P(B |A) if A and B are dependent |
The conditional probability of B given A is the probability of event B occurring, given that event A has already occurred. Using the dependent equation above and then dividing by P(A) we have a formula for conditional probability.
EXAMPLE:
Two cards are drawn from a well-shuffled deck of cards. What is the probability of drawing an ace and a non-ace if
(a) the first card is replaced and the deck is shuffled before the second is drawn, and
(b) the cards are drawn consecutively without replacement?
SOLUTION:
(a) Independent events. P(ace and non-ace) = P(ace) • P(non-ace) = 4/52 x 48/52 = 192/2704 = 12/169
(b) Dependent events. P(ace and non-ace) = P(ace) • P(non-ace | ace) = 4/52 x 48/51 = 192/2652 = 16/221
The previous pages are a basic look into probability. There are many more topics that could be explored. These topics include probability involving combinations and permutations, binominal distribution, expected values, and odds.
The importance of probability is continuing to grow as it is used by more and more political scientists, economists, biologists, actuaries, business executives and other professionals.