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

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Question

In the carnival game Under-or-Over-Seven, a pair of tair dice is rolled once, and the resulting sum determines whether the example, using method one, the player can bet $3.50 that the sum will be under 7, that is player wins or loses his or her bet. For , 2, 3, 4, 5, or 6. For this bet, the player wins $3.50 if the result is under 7 and outcome equals or is greater than 7. Similarly, using method two, the player can bet $3.50 that the sum will be over 7, that is, 8, 9, 10, 11, or 12. Here, if the result is 7 or under. A third method of play is to bet $3.50 on the outcome 7. For this bet, the player wins the player wins $3.50 if the result is over 7 but loses $3.50 $14.00 if the result of the roll is 7 and loses $3.50 otherwise. Complete parts (a) through (d) E Click the icon to view a table of all possible outcomes of a two dice roll. a. Construct the probability distibution representing the different outcomes that are possible for a $3.50 bet using method one P(X) $3.5 15 36 -$3.5 21 36 Type an exact answer in simplified form.) b. Construct the probability distribution representing the different outcomes that are possible for a $3.50 bet using method two Type an exact answer in simplified form.)

Explanation / Answer

b) here as probability of 7 or over =15/36

and probability of equal to or less than 7 =21/36

hence

x P(X) 3.5 15/36 -3.5 21/36

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

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