Before we can compute any probabilities for outcomes in a random process, we have to define the sample space , or collection of all possible outcomes. If we have listed all outcomes and it is reasonable to assume that the outcomes are

Let's consider a simplified lottery game. Suppose that Ohio has a game where you try to guess a random two-digit number that is selected. This "winning" random number is selected by the following process. There are two boxes, labelled box A and box B, that each contain 10 ping pong-balls labelled using the digits 0 through 9. A random number is selected by letting the first digit be the number of the ball selected from box A, and the second digit is the number of the ball selected from box B.

What is the sample space? There are 100 possible winning two-digit numbers that are listed below.

00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99

By the way the two-digit number is selected, no particular number listed above has any more or less chance of being selected than another number. So it is reasonable to assign the same probability to each number in the above list. What probability should be assigned? There are 100 possible winning numbers. If we wish to assign the same probability to each number and keep the total probability of all the numbers equal to 1, then each number should be given the probability 1/100 = .01.

In general, if there are * N * possible outcomes in an experiment
and the outcomes are * equally likely *, then you should assign
a probability of * 1/N *to each outcome.

**
CAUTION:
** This recipe for assigning probabilities works only when the outcomes
are equally likely. It is easy to misuse this. For example, suppose you
toss a coin three times and you're interested in the number of heads.
The possible numbers of heads (the sample space) are

0 head, 1 head, 2 heads, 3 heads

There are four outcomes in this case. But it is incorrect to assume that the
probabilities of each outcome is 1/4 = .25. These four outcomes are *
not * equally likely. In fact, the probability of 1 head is three times
the probability of 3 heads.

Return to AN INTRODUCTION TO PROBABILITY

Page Author: Jim Albert (© 1996)

albert@bayes.bgsu.edu

Document: http://www-math.bgsu.edu/~albert/m115/probability/equally_likely.html

Last Modified: November 18, 1996