Answer to Question #169910 in Statistics and Probability for Rustom

Question #169910

Construct the probability distribution for the random variables described in

each of the following situations.

1. Two dice are tossed. Let X = represent the sum of two dice

Make a table of all possible values. Find the values of the random

variable X.

Value of the Random Variable X

Probability P(X)


1
Expert's answer
2021-03-10T07:12:04-0500

Find the probability that the sum is 2:

"P(2) = \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{1}{{36}}" ((1,1) - the first and second dice rolled one point each)

Find the probability that the sum is 3:

"P(3) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 = \\frac{2}{{36}} = \\frac{1}{{18}}" ((1,2) or (2,1) - 1 point was dropped on the first dice, and 2 points on the second, or vice versa)

Find the probability that the sum is 4:

"P(4) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{3}{{36}} = \\frac{1}{{12}}" ((1,3) or (3,1) or (2,2) - the first dice dropped 1 point, and the second 3 points, or vice versa, or both dice dropped 2 points)

Find the probability that the sum is 5:

"P(5) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 = \\frac{4}{{36}} = \\frac{1}{{9}}" ((1,4) or (4,1) or (2,3) or (3,2))

Find the probability that the sum is 6:

"P(6) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{5}{{36}}" ((1,5), (5,1), (2,4), (4,2), (3,3))

Find the probability that the sum is 7:

"P(7) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 = \\frac{6}{{36}} = \\frac{1}{6}" ((1,6) or (6,1) or (2,5) or (5,2) or (3,4) or (4,3))

Find the probability that the sum is 8:

"P(8) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{5}{{36}}" ((2,6) or (6,2) or (5,3) or (3,5) or (4,4))

Find the probability that the sum is 9:

"P(9) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 = \\frac{4}{{36}} = \\frac{1}{9}" ((3,6) or (6,3) or (5,4) or (4,5))

Find the probability that the sum is 10:

"P(10) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 + \\frac{1}{6} \\cdot \\frac{1}{6} = \\frac{3}{{36}} = \\frac{1}{{12}}" ((4,6) or (6,4) or (5,5))

Find the probability that the sum is 11:

"P(11) = \\frac{1}{6} \\cdot \\frac{1}{6} \\cdot 2 = \\frac{2}{{36}} = \\frac{1}{{18}}" ((5,6) or (6,5) )

Find the probability that the sum is 12:

"P(12) = \\frac{1}{6} \\cdot \\frac{1}{6} \\ = \\frac{1}{{36}}" (6,6 )

We have the probability distribution:

"\\begin{matrix}\nX&2&3&4&5&6&7&8&9&{10}&{11}&{12}\\\\\np&{\\frac{1}{{36}}}&{\\frac{1}{{18}}}&{\\frac{1}{{12}}}&{\\frac{1}{9}}&{\\frac{5}{{36}}}&{\\frac{1}{6}}&{\\frac{5}{{36}}}&{\\frac{1}{9}}&{\\frac{1}{{12}}}&{\\frac{1}{{18}}}&{\\frac{1}{{36}}}\n\\end{matrix}"





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