Agame of chance involves rolling an unevenly balanced 4-sided die. the probability that a roll comes up 1 is 0.21, the probability that a roll comes up 1 or 2 is 0.42, and the probability that a roll comes up 2 or 3 is 0.51 . if you win the amount that appears on the die, what is your expected winnings? (note that the die has 4 sides.) answer = dollars.

2.65 dollars

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

E(x)=x1P(x1) + x2P(x2) + ... + xnP(xn)

Where x1, x2, ... , xn are the possibles values of the variable and P(x1), P(x2), ... , P(xn) are the probabilities of x1, x2, ... , xn respectively.

In this case, the roll can comes up 1, 2, 3 or 4 and you can win 1, 2, 3 or 4 dollars respectively. So, taking into account that they are mutually exclusive events, the probability that the player win 1, 2, 3 or 4 dollars is:

P(1) = 0.21

P(2) = P(1∪2) - P(1) = 0.42 - 0.21 = 0.21

P(3) = P(2∪3) - P(2) = 0.51 - 0.21 = 0.3

P(4) = 1 - P(1) - P(2) - P(3) = 1 - 0.21 - 0.21 - 0.3 = 0.28

Therefore, If you win the amount that appears on the die the expected winning are:

E(x) = $1P(1) + $2P(2) + $3P(3) + $4P(4)

E(x) = $1(0.21) + $2(0.21) + $3(0.3) + $4(0.28)

E(x) = $2.65

remove the radical and try not to simplify

Step-by-step explanation: