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?

Answer 1

It's easy enough. Solving looks like that: p(roll of 8)+p(roll of 12) =$\frac{5}{36} + \frac{1}{36} = \frac{1}{6} ; \frac{1}{6} *9 +\frac{5}{6}*2 = \frac{3}{2} - \frac{5}{3} = -\frac{1}{6}$ Hope everything is clear.

Answer 2

Answer:  $-\frac{1}{6}$

 

Step-by-step explanation:

Since, the total outcomes when two dices are thrown = 36

Also, the total outcomes of getting the sum of 8 or 12 in throwing = {(2,6)(6,2)(3,5),(5,3),(4,4)}= 6

Hence, the probability of getting the sum of 8 or 12 in throwing two dices

$= \frac{6}{36}=\frac{1}{6}$

Also,  the probability of getting the sum of 8 or 12 in throwing two dices $= 1-\frac{1}{6}=\frac{5}{6}$

Thus, the expected point he can earn in one throw $= 9\times \frac{1}{6}-2\times {5}{6} = \frac{9}{6}-\frac{10}{6}=-\frac{1}{6}$