Craps Game Probability?

I don't usually ask for help on math problems but I am truly stumped on this one.

In a game of craps, you roll two fair dice. Whether you win or lose depends on the sum of the numbers occurring on the tops of the dice. Let x be the random variable that represents the sum of the numbers on the tops of the dice.

What is the probability distribution of these x values (that is, what is the probability that x = 2, 3, etc.)? (Round your answers to three decimal places.)

Probability of x=2?
Probability of x=3?
Probability of x=4?
Probability of x=5?
Probability of x=6?
Probability of x=7?
Probability of x=8?
Probability of x=9?
Probability of x=10?
Probability of x=11?
Probability of x=12?

Anonymous2011-11-02T22:00:23Z

Favorite Answer

x=2, 1/36 or .028
x=3, 1/18 or .056
x=4, 1/12 or .083
x=5, 1/9 or .111
x=6, 5/36 or .139
x=7, 1/6 or .167
x=8, 5/36 or .139
x=9, 1/9 or .111
x=10, 1/12 or .083
x=11, 1/18 or .056
x=12, 1/36 or .028

11

?2017-01-17T08:38:57Z

Thats a soreness interior the dick you may parent the probibility of no longer triumphing for each factor, and then evaluate how in many circumstances each and each factor is the factor in accordance with its probibility of rolling in the process the pop out roll. You cant confirm the probibility based on the Nth roll era, because of the fact the sport transformations after the 1st comeout, you could parent a fastened probibility of triumphing for the 1st roll, and then use a variable for all next rolls. in case you paypal me 50 funds unwell parent it out for you and supply you very particular counsel.

00

?2011-11-02T22:04:41Z

There are only 36 numbers possible.

One 2 (1,1) = 1 out of 36

two 3s (1,2) (2,1) = 2 out of 36

three 4s ( 1,3) (2,2) (3,!)

four 5s (1,4) (2,3) (3,2) (4,1)

five 6s (1,5) (2,4) (3,3) (4,2) (5,1)

six 7s ( 1,6) (2,4) (3,4) (4,3) ( 5,2) (6,1)

five 8s ( 6,2) (5,3) (4,4) (3,5) (2,6) = 5 out of 36

four 9s

three 10s

two 11s

one 12 (6,6)

02

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