Question

harlene tosses two number cubes. if a sum of 8 or 12 comes up, she gets 9 points. if not, she loses 2 points. what is the expected value of the number of points for one roll?
$-\frac{2}{3}$
$-\frac{1}{6}$
$\frac{1}{6}$
$\frac{2}{3}$

Explanation:

Step1: Find total outcomes

When tossing two number cubes (each with 6 faces), total outcomes = \(6\times6 = 36\).

Step2: Find favorable outcomes for sum 8 or 12

• Sum of 8: Possible pairs \((2,6),(3,5),(4,4),(5,3),(6,2)\) → 5 outcomes.
• Sum of 12: Only \((6,6)\) → 1 outcome.
• Total favorable for 8 or 12: \(5 + 1 = 6\) outcomes.

Step3: Calculate probabilities

• Probability of sum 8 or 12: \(P(\text{success})=\frac{6}{36}=\frac{1}{6}\).
• Probability of not sum 8 or 12: \(P(\text{failure}) = 1-\frac{1}{6}=\frac{5}{6}\).

Step4: Calculate expected value

Expected value \(E = (\text{Points for success} \times P(\text{success})) + (\text{Points for failure} \times P(\text{failure}))\)
\(E = 9\times\frac{1}{6}+(-2)\times\frac{5}{6}=\frac{9}{6}-\frac{10}{6}=\frac{9 - 10}{6}=-\frac{1}{6}\).

Answer:

\(-\frac{1}{6}\) (corresponding to the option "-\(\frac{1}{6}\)")