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the carnival game Under-or-Over-Seven, a pair of fair dice is rolled once, and the resulting sum...

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, the player can bet $1 that the sum will be under 7—that is, 2, 3, 4, 5, or 6. For this bet, the player wins $1 if the result is under 7 and loses $1 if the outcome equals or is greater than 7. Similarly, the player can bet $1 that the sum will be over 7—that is, 8, 9, 10, 11, or 12. Here, the player wins $1 if the result is over 7 but loses $1 if the result is 7 or under. A third method of play is to bet $1 on the outcome 7. For this bet, the player wins $4 if the result of the roll is 7 and loses $1 otherwise.

(a) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on being under 7.

(b) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on being over 7.

(c) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on 7.

math   Statistics-And-Probability   
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Answer #1

X : resulting sum when two fair dice rolled

When total dice are rolled there are a total 36 (6x6) outcomes . Each outcome with equal probability of 1/36

Following is sample space for different values of X:Sum

X : 2 : (1,1) ; P(X=2)=1/36

X : 3 : (1,2) (2,1): P(X=3) =2/36

X : 4 : (1,3)(2,2)(3,1) ; P(X=4)=3/36

X : 5 : (1,4)(2,3)(3,2)(4,1) ; P(X=5)=4/36

X : 6 : (1,5) (2,4)(3,3)(4,2)(4,1): P(X=6) =5/36

X : 7 : (1,6)(2,5)(3,4)(4,3)(5,2)(6,1) ; P(X=7)=6/36

X : 8 : (2,6) (3,5)(4,3)(5,3)(6,2): P(X=8) =5/36

X: : 9 :(3,6)(4,5)(5,4)(6,3) : P(X=9) = 4/36

X : 10 : (4,6)(5,5)(6,4) : P(X=10) = 3/36

X : 11 : (5,6)(6,5) : P(X=11) = 2/36

X: 12 : (6,6) : P(X=12)

(a) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on being under 7.

player wins $1 if the result is under 7

W : Winning amount

player wins $1 if the result is under 7 Winning the bet $1 i.e W = 1 if P(Sum < 7) =P(X<7)

player loses $1 otherwise i.e result is 7i.e W = - 1 if P(Sum 7) = 1-P(Sum < 7)

P(Sum < 7) = P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 = 15/36

P(Sum 7) = 1-P(Sum < 7) = 1-15/36 = 21/36

probability distribution representing the different outcomes that are possible for a $1 bet on being under 7

W:Winning P(w) P(w)
$1 15/36 0.416666667
-$1 21/36 0.583333333

(b) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on being over 7.

player wins $1 if the result is over 7

W : Winning amount

player wins $1 if the result is over 7 Winning the bet $1 i.e W = 1 if P(Sum > 7) =P(X>7)

player loses $1 otherwise i.e result is 7i.e W = - 1 if P(Sum 7) = 1-P(Sum > 7)

P(Sum > 7) = P(X=8)+P(X=9)+P(X=10)+P(X=11)+P(X=12) = 1/36 + 2/36 + 3/36 + 4/36 + 5/36 = 15/36

P(Sum 7) = 1-P(Sum > 7) = 1-15/36 = 21/36

probability distribution representing the different outcomes that are possible for a $1 bet on being over 7

W:Winning P(w) P(w)
$1 15/36 0.416666667
-$1 21/36 0.583333333

(c) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on 7.

player wins $1 if the result is 7

W : Winning amount

player wins $1 if the result is 7 Winning the bet $1 i.e W = 1 if P(Sum = 7) =P(X=7)

player loses $1 otherwise i.e result is not equal to  7i.e W = - 1 if P(Sum 7) = 1-P(Sum = 7)

P(Sum = 7) = 6/36 = 1/6

P(Sum 7) = 1-P(Sum = 7) = 1-1/6 = 5/6

probability distribution representing the different outcomes that are possible for a $1 bet on 7

W:Winning P(w) P(w)
$1 1/6 0.166666667
-$1 5/6 0.833333333
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