Section 9.1 Probability Distributions and Expected
Value
Random Variables
One of the questions asked in the 2010 National Health and Nutrition Examination Study
(NHANES) had to do with respondents’ daily hours of TV or video use. The answer to that
question, which we will label x, is one of the numbers 0 through 6 (corresponding to the Q
numbers of hours of use). Since the value of x is random, x is called a random variable.
The following table gives each possible outcome of the study question on TV and video use together with the probability P (x) of each outcome x. (Data from: www.cdc.gov/nchs/nhanes.htm.)
x
P (x)
0
.13
1
.24
2
.33
3
.14
4
.07
5
.07
6
.02
A table that lists all the outcomes with the corresponding probabilities is called a probability
distribution. The sum of the probabilities in a probability distribution must always equal 1.
(The sum in some distributions may vary slightly from 1 because of rounding.)
The information in a probability distribution is often displayed graphically as a special kind
of bar graph called a histogram. The bars of a histogram all have the same width, usually 1
unit. The heights of the bars are determined by the probabilities. A histogram for the data in
the probability distribution presented above is given in the Figure below (left). A histogram
shows important characteristics of a distribution that may not be readily apparent in tabular
form, such as the relative sizes of the probabilities and any symmetry in the distribution.
The area of the bar above x = 0 in the Figure above (left) is the product of 1 and .13 , or
1·.13 = .13. Since each bar has a width of 1, its area is equal to the probability that corresponds
to its x-value. The probability that a particular value will occur is thus given by the area of the
appropriate bar of the graph. For example, the probability that one or more hours are spent
watching TV or a video is the sum of the areas for x = 1, x = 2, x = 3, x = 4, x = 5, and
1
x = 6. This area, shown in red in the Figure above (right), corresponds to .87 of the total area,
since
P (x ≥ 1) = P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) + P (x = 6)
= .24 + .33 + .14 + .07 + .07 + .02
= .87.
EXAMPLE:
(a) Give the probability distribution for the number of heads showing when two coins are tossed.
Solution: Let x represent the random variable “number of heads.” Then x can take on the value
0, 1, or 2. Now find the probability of each outcome. When two coins are tossed, the sample
space is {T T, T H, HT, HH}. So the probability of getting one head is 2/4 = 1/2. Similar
analysis of the other cases produces this table.
x
P (x)
0
1/4
1
1/2
2
1/4
(b) Draw a histogram for the distribution in the table. Find the probability that at least one
coin comes up heads.
Solution: The histogram is shown in the Figure below. The portion in red represents
P (x ≥ 1) = P (x = 1) + P (x = 2) =
3
1 1
+ =
2 4
4
Expected Value
In working with probability distributions, it is useful to have a concept of the typical or average
value that the random variable takes on. In the Example above, for instance, it seems reasonable
that, on the average, one head shows when two coins are tossed. This does not tell what will
happen the next time we toss two coins; we may get two heads, or we may get none. If we
tossed two coins many times, however, we would expect that, in the long run, we would average
about one head for each toss of two coins.
A way to solve such problems in general is to imagine flipping two coins 4 times. Based on the
probability distribution in the Example above, we would expect that 1 of the 4 times we would
get 0 heads, 2 of the 4 times we would get 1 head, and 1 of the 4 times we would get 2 heads.
The total number of heads we would get, then, is
0·1+1·2+2·1=4
2
The expected number of heads per toss is found by dividing the total number of heads by the
total number of tosses:
1
1
1
0·1+1·2+2·1
=0· +1· +2· =1
4
4
2
4
Notice that the expected number of heads turns out to be the sum of the three values of the
random variable x, multiplied by their corresponding probabilities. We can use this idea to
define the expected value of a random variable as follows.
EXAMPLE: Consider a lottery with a single JACKPOT prize of $500,000. If a ticket costs $3,
and the probability of it being a winning ticket is approximately 1/1,000,000, then what is the
expected winnings?
Solution: We have
Event
win
lose
Winnings
$500,000-$3
-$3
Probability
1/1,000,000
999,999/1,000,000
The expected winnings is
999, 999
1
+ (−$3)
= −$2.50
1, 000, 000
1, 000, 000
EXAMPLE: A contractor is bidding on a job that promises a profit of $200,000 with a probability of 7/10 and a loss, due to strikes, weather conditions, late arrival of building materials,
and so on, of $40,000 with a probability of 3/10. Determine the contractor’s expected profit.
($499, 997)
Solution: We have
3
7
+ (−$40, 000) = $128, 000
10
10
EXAMPLE: Each day, Lynette and Tanisha toss a coin to see who buys coffee (at $1.75 a cup).
One tosses, while the other calls the outcome. If the person who calls the outcome is correct,
the other buys the coffee; otherwise the caller pays. Find Lynette’s expected winnings.
($200, 000)
Solution: Assume that an honest co in is used, that Tanisha tosses the coin, and that Lynette
calls the outcome. The possible results and corresponding probabilities are shown in the following table:
Lynette wins a $1.75 cup of coffee whenever the results and calls match, and she loses $1.75
when there is no match. Her expected winnings are
1
1
1
1
+ (−1.75)
+ (−1.75)
+ 1.75
=0
1.75
4
4
4
4
On the average, over the long run, Lynette breaks even.
3