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) = 1/36 p(3) = 1/18 p(4) = 1/12 p(5) = 1/9 p(6) = 5/36 p(7) = 1/6 p(8) = 5/36 p(9) = 1/9 p(10) = 1/12 p(11) = 1/18 p(12) = 1/36 the probability of a sum of 7 or 11 is: 1/36 1/18 2/9 1/36 done

Explanation:

Step1: Identify relevant probabilities

We know $P(7)=\frac{1}{6}$ and $P(11)=\frac{1}{18}$.

Step2: Use addition - rule for mutually - exclusive events

Since getting a sum of 7 and getting a sum of 11 are mutually - exclusive events, we use $P(A\cup B)=P(A)+P(B)$. So $P(7\ or\ 11)=P(7)+P(11)$.

Step3: Calculate the sum

$P(7)+P(11)=\frac{1}{6}+\frac{1}{18}=\frac{3 + 1}{18}=\frac{4}{18}=\frac{2}{9}$.

Answer:

$\frac{2}{9}$