P - Probability Distributions Lesson

Math_Lesson_TopBanner.png 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}P(X=2)=136

3

2

LaTeX: P\left(X=3\right)=\frac{2}{36}=\frac{1}{18}P(X=3)=236=118

4

3

LaTeX: P\left(X=4\right)=\frac{3}{36}=\frac{1}{12}P(X=4)=336=112

5

4

LaTeX: P\left(X=5\right)=\frac{4}{36}=\frac{1}{9}P(X=5)=436=19

6

5

LaTeX: P\left(X=6\right)=\frac{5}{36}P(X=6)=536

7

6

LaTeX: P\left(X=7\right)=\frac{6}{36}=\frac{1}{6}P(X=7)=636=16

8

5

LaTeX: P\left(X=8\right)=\frac{5}{36}P(X=8)=536

9

4

LaTeX: P\left(X=9\right)=\frac{4}{36}=\frac{1}{9}P(X=9)=436=19

10

3

LaTeX: P\left(X=10\right)=\frac{3}{36}=\frac{1}{12}P(X=10)=336=112

11

2

LaTeX: P\left(X=11\right)=\frac{2}{36}=\frac{1}{18}P(X=11)=236=118

12

1

LaTeX: P\left(X=12\right)=\frac{1}{36}P(X=12)=136

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.

Histogram.png

We can also calculate the expected value:

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)=7E(x)=2(136)+3(118)+4(112)+5(19)+6(536)+7(16)+8(536)+9(19)+10(112)+11(118)+12(136)=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}\approx0.189500.18 - 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}\approx0.145360.14

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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