One option in a roulette game is to bet $12 on red. (there are 18 red compartments, 18 black compartments, and two compartments that are neither red nor black.) if the ball lands on red, you get to keep the $ 12 you paid to play the game and you are awarded $ 12. if the ball lands elsewhere, you are awarded nothing and the $ 12 that you bet is collected. complete parts (a) through (b) below. a. what is the expected value for playing roulette if you bet $12 on red?

The expected value is -$0,63

Step-by-step explanation:

The expected value of a discrete variable x is given by:

E(x)=x1*p(x1)+x2*p(x2)+...+xn*p(xn)

Where x1, x2, .. xn are the posibles results that the variable x can take and p(x1), p(x2),...,P(xn) are their respective probabilities.

In this case, we have two possible results if we bet $12 on red. This results are:

1. The balls lands on red: in this case we keep the $12 we paid and we awarded 12. That means, we gain $12 and the probability of this case is the division of the number of red compartments by total number of compartments. So:

x1=$12

p(x1)=18/38

2. The balls doesn't land on red: in this case we lose the $12 that we bet and the probability is the division of number of compartments that aren't red by the total number of compartments. So:

x2=-$12

p(x2)=20/38

Then, replacing x1, x2, p(x1) and p(x2) on the equation of the expected value, we get:

E(x)=($12)*(18/38)  +  (-$12)*(20/38)

E(x)=$-0.63

So, the expected value if we bet $12 on red is to lose $0.63

the solution is 12x-6y

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation: