Answer to Question #306738 in Statistics and Probability for tinny

Question

4. Using the sample space for rolling two dice, construct a probability distribution for the random variable X representing the sum of the numbers that appear.

Answer 1

1. 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

2. Probability that the sum is 3

$P(3) = \frac{1}{6} \cdot \frac{1}{6} \cdot 2 = \frac{2}{{36}}$

(1,2) or (2,1) - 1 point was rolled on the first dice, and 2 points on the second, or vice versa

3. 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 rolled 1 point, and the second 3 points, or vice versa, or both dice rolled 2 points

4. 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)

5. 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)

6. 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)

7. 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)

8. 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)

9. 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)

10. 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)

11. Probability that the sum is 12

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

(6,6 )

The Probability Distribution

$\begin{matrix} X&2&3&4&5&6&7&8&9&{10}&{11}&{12}\\ p&{\frac{1}{{36}}}&{\frac{1}{{18}}}&{\frac{1}{{12}}}&{\frac{1}{9}}&{\frac{5}{{36}}}&{\frac{1}{6}}&{\frac{5}{{36}}}&{\frac{1}{9}}&{\frac{1}{{12}}}&{\frac{1}{{18}}}&{\frac{1}{{36}}} \end{matrix}$