Question

you toss two number cubes. if a sum of 7 or 11 comes up, you get 7 points. if not you lose 2 points.
the probabilities for each of the sums is:
$p(2)=\frac{1}{36}quad p(3)=\frac{1}{18}quad p(4)=\frac{1}{12}quad p(5)=\frac{1}{9}$
$p(6)=\frac{5}{36}quad p(7)=\frac{1}{6}quad p(8)=\frac{5}{36}quad p(9)=\frac{1}{9}$
$p(10)=\frac{1}{12}quad p(11)=\frac{1}{18}quad p(12)=\frac{1}{36}$
the probability of a sum of 7 or 11 is
the expected value of the number of points for one roll is

Explanation:

Step1: Calculate the probability of sum 7 or 11

We know that \(P(7)=\frac{1}{6}\) and \(P(11)=\frac{1}{18}\).
Using the formula for the probability of the union of two mutually - exclusive events \(P(A\cup B)=P(A)+P(B)\) (since getting a sum of 7 and getting a sum of 11 when rolling two dice are mutually - exclusive events).
\(P(7\cup11)=P(7)+P(11)=\frac{1}{6}+\frac{1}{18}\)
First, find a common denominator, which is 18. Then \(\frac{1}{6}=\frac{3}{18}\).
So \(P(7\cup11)=\frac{3 + 1}{18}=\frac{4}{18}=\frac{2}{9}\)

Step2: Calculate the expected value

Let \(X\) be the random variable representing the number of points.
If the sum is 7 or 11 (\(P(X = 7)=\frac{2}{9}\)), and if the sum is not 7 or 11 (\(P(X=-2)=1 - \frac{2}{9}=\frac{7}{9}\))
The formula for the expected value \(E(X)=\sum_{i}x_ip_i\)
\(E(X)=7\times\frac{2}{9}+(- 2)\times\frac{7}{9}\)
\(E(X)=\frac{14}{9}-\frac{14}{9}=0\)

Answer:

The probability of a sum of 7 or 11 is \(\frac{2}{9}\). The expected value of the number of points for one roll is \(0\).