Chapter 5.2, Problem 58E

To determine

Probability of getting craps and or an even sum on one roll of the dice.

Answer to Problem 58E

Probability,

  $P\left(\text{C}\cup \text{E}\right)=\frac{20}{36}$

Explanation of Solution

Given information:

Assignment of probabilities to the sum of the numbers on the up − faces when two dice are rolled:

  

Calculations:

Two Events:

C: Craps

E: Even sum

Let the sum of outcomes of two dice be x .

Probabilities of outcomes are as follows:

For x = 2:

  $P\left(x=2\right)=\frac{1}{36}$

For x = 3:

  $P\left(x=3\right)=\frac{2}{36}$

For x = 4:

  $P\left(x=4\right)=\frac{3}{36}$

For x = 5:

  $P\left(x=5\right)=\frac{4}{36}$

For x = 6:

  $P\left(x=6\right)=\frac{5}{36}$

For x = 7:

  $P\left(x=7\right)=\frac{6}{36}$

For x = 8:

  $P\left(x=8\right)=\frac{5}{36}$

For x = 9:

  $P\left(x=9\right)=\frac{4}{36}$

For x = 10:

  $P\left(x=10\right)=\frac{3}{36}$

For x = 11:

  $P\left(x=11\right)=\frac{2}{36}$

For x = 12:

  $P\left(x=12\right)=\frac{1}{36}$

When a player rolls a 2, 3, or 12, he gets craps.

Apply addition rule of mutually exclusive events to add the corresponding probabilities:

  $\begin{array}{l}P\left(\text{C}\right)=P\left(x=2\right)+P\left(x=3\right)+P\left(x=12\right)\\ =\frac{1}{36}+\frac{2}{36}+\frac{1}{36}\\ =\frac{4}{36}\end{array}$

When a player rollsan even sum, he gets 2, 4, 6, 8, 10 or 12.

Apply addition rule of mutually exclusive events to add the corresponding probabilities:

  $\begin{array}{l}P\left(\text{E}\right)=P\left(x=2\right)+P\left(x=4\right)+P\left(x=6\right)+P\left(x=8\right)+P\left(x=10\right)+P\left(x=12\right)\\ =\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36}\\ =\frac{18}{36}\end{array}$

When a player rolls craps and an even sum, he gets 2 or 12.

Apply addition rule of mutually exclusive events to add the corresponding probabilities:

  $\begin{array}{l}P\left(\text{C}\cap \text{E}\right)=P\left(x=2\right)+P\left(x=12\right)\\ =\frac{1}{36}+\frac{1}{36}\\ =\frac{2}{36}\end{array}$

Finally,

Apply general addition rule for any two events:

  $\begin{array}{l}P\left(\text{C}\cup \text{E}\right)=P\left(\text{C}\right)+P\left(\text{E}\right)-P\left(\text{C}\cap \text{E}\right)\\ =\frac{4}{36}+\frac{18}{36}-\frac{2}{36}\\ =\frac{20}{36}\end{array}$