Probability Distributions

We can use theoretical probability to create a probability distribution for a certain situation. A theoretical probability is a probability created from what we know about a situation (without conducting an experiment). When we use theoretical probability, we often make a table of all of the possible outcomes and then analyze those possibilities.

Let's assign the random variable X to be the sum after rolling two dice. A random variable is used for probabilities (it is different than using the letter x to represent some number!). Let's go ahead and determine all of the possible outcomes.

	1	2	3	4	5	6
1	(1, 1)	(1, 2)	(1, 3)	(1, 4)	(1, 5)	(1, 6)
2	(2, 1)	(2, 2)	(2, 3)	(2, 4)	(2, 5)	(2, 6)
3	(3, 1)	(3, 2)	(3, 3)	(3, 4)	(3, 5)	(3, 6)
4	(4, 1)	(4, 2)	(4, 3)	(4, 4)	(4, 5)	(4, 6)
5	(5, 1)	(5, 2)	(5, 3)	(5, 4)	(5, 5)	(5, 6)
6	(6, 1)	(6, 2)	(6, 3)	(6, 4)	(6, 5)	(6, 6)

So, there are 36 different possible outcomes. Let's create a probability distribution for all of the different possibilities for X. The possible sums could be 2 to 12.

X = sum of two rolls	Number of outcomes that satisfy that sum	Probability
2	1	$LaTeX: P\left(X=2\right)=\frac{1}{36}$
3	2	$LaTeX: P\left(X=3\right)=\frac{2}{36}=\frac{1}{18}$
4	3	$LaTeX: P\left(X=4\right)=\frac{3}{36}=\frac{1}{12}$
5	4	$LaTeX: P\left(X=5\right)=\frac{4}{36}=\frac{1}{9}$
6	5	$LaTeX: P\left(X=6\right)=\frac{5}{36}$
7	6	$LaTeX: P\left(X=7\right)=\frac{6}{36}=\frac{1}{6}$
8	5	$LaTeX: P\left(X=8\right)=\frac{5}{36}$
9	4	$LaTeX: P\left(X=9\right)=\frac{4}{36}=\frac{1}{9}$
10	3	$LaTeX: P\left(X=10\right)=\frac{3}{36}=\frac{1}{12}$
11	2	$LaTeX: P\left(X=11\right)=\frac{2}{36}=\frac{1}{18}$
12	1	$LaTeX: P\left(X=12\right)=\frac{1}{36}$

Notice: All of these probabilities add to 1!!!

We can graph this probability distribution using a histogram, we'd put each outcome on the x-axis and the probability on the y-axis.

We can also calculate the expected value:

equation image indicator $LaTeX: E\left(x\right)=2\left(\frac{1}{36}\right)+3\left(\frac{1}{18}\right)+4\left(\frac{1}{12}\right)+5\left(\frac{1}{9}\right)+6\left(\frac{5}{36}\right)+7\left(\frac{1}{6}\right)+8\left(\frac{5}{36}\right)+9\left(\frac{1}{9}\right)+10\left(\frac{1}{12}\right)+11\left(\frac{1}{18}\right)+12\left(\frac{1}{36}\right)=7$

We can also conduct an experiment to consider the empirical probability. The empirical probability uses an experiment to determine the probability of a situation happening.

Below is a table of the results from an experiment when two dice were rolled 50 times.

Sum of Rolls of Two Dice
4, 6, 4, 8 , 10, 8 , 6, 11, 9, 2, 3, 6, 12, 5, 6, 8 , 4, 8 , 7, 11, 5, 2, 8 , 3, 9, 4, 10, 3, 5, 3, 7, 8 , 3, 11, 6, 4, 8 , 3, 6, 4, 11, 9, 7, 8 , 5, 9, 3, 6, 8 , 4

a. Using the table, what is the empirical probability of rolling a sum of 8?

Solution: $LaTeX: \frac{9}{50}\approx0.18$ - because there were nine times that the rolls summed to 8 out of 50 experiments

b. Using the probability distribution above, what is the theoretical probability of rolling a sum of 8?

Solution: $LaTeX: \frac{5}{36}\approx0.14$

c. How do the empirical and theoretical probabilities compare?

Solution: The rolls summed to 8 more times than expected in the experiment. We would expect there to be 0.14(50)=6.9 rolls that summed to 8 out of the 50 rolls.

Probability Distribution Practice

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