In the carnival game Under-or-Over-Seven, a pair of fair dice is rolled once, an

Question

In the carnival game Under-or-Over-Seven, a pair of fair dice is rolled once, and the resulting sum determines whether the player wins or loses his or her bet. For example, using method one, the player can bet $2.00 that the sum will be under 7, that is, 2, 3, 4, 5, or 6. For this bet, the player wins $2.00 if the result is under 7 and loses $2.00 if the outcome equals or is greater than 7. Similarly, using method two the player can be $200 that hes will be o eT a i 10. 1 or12 le the player wins $2.00 if the result is over 7 but loses $2.00 if the result is 7 or under. A third method of play is to bet $2.00 on the outcome 7. For this bet, the player wins $8.00 if the result of the roll is 7 and loses $2.00 otherwise. Complete parts (a) through (d). Click the icon to view a table of all possible outcomes of a two dice roll. a. Construct the probability distribution representing the different outcomes that are possible for a $2.00 bet using method one $2 7 -$ 2 12 Type an exact answer in simplified form.) b. Construct the probability distribution representing the different outcomes that are possible for a $2.00 bet using method two. P(X) Type an exact answer in simplified form.)

Explanation / Answer

a. There are 36 outcomes one would get for rolling a pair of fair dice. For sum to be under 7, the possible outcomes are, 2, 3, 4, 5 and 6. The probabilities of the outcomes are 1/36, 2/36, 3/36, 4/36, 5/36. Assume, the r.v denote the sum of the outcom eof dice. Therefore, the P(X<=6)=1/36+2/36+...+5/36=0.4167

The sum of 7 or more are 7, 8, 9, 10, 11 and 12. Thus, P(X>=7)=6/36+5/36+4/36+3/36+2/36+1/36=0.5833

P($2)=0.4167

P(-$2)=0.5833

b. P($2)=P(X>7)=5/36+4/36+3/36+2/36+1/36=0.4167

P(-$2)=P(X<=7)=1/36+2/36+3/36+4/36+5/36+6/36=0.5833

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