91Ó°ÊÓ

When rolling two dice, their numbers are picked from the set {1, 2, 3, 4, 5, 6}. The random variable X is the smaller number of the two dice. The possible outcomes for X are the numbers 1 to 6. To determine the probability of each value of X, we can count the number of ways each value could appear as the smaller number, and then divide this count by the total number of outcomes of rolling two dice, which is 36.

For each value of X, we will calculate the probability of X being that value by counting the number of outcomes that result in that value and dividing it by the total number of outcomes (which is 36, as there are 6x6 outcomes when rolling two dice). \(P(X=1)\): There are 6 outcomes where the lower number is 1: (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) \( P(X=1)=\dfrac{6}{36} \) \(P(X=2)\): There are 5 outcomes where the lower number is 2: (2,2), (2,3), (2,4), (2,5), (2,6) \( P(X=2)=\dfrac{5}{36} \) \(P(X=3)\): There are 4 outcomes where the lower number is 3: (3,3), (3,4), (3,5), (3,6) \( P(X=3)=\dfrac{4}{36} \) \(P(X=4)\): There are 3 outcomes where the lower number is 4: (4,4), (4,5), (4,6) \( P(X=4)=\dfrac{3}{36} \) \(P(X=5)\): There are 2 outcomes where the lower number is 5: (5,5), (5,6) \( P(X=5)=\dfrac{2}{36} \) \(P(X=6)\): There is only 1 outcome where the lower number is 6: (6,6) \( P(X=6)=\dfrac{1}{36} \)

To find the expected value of X, we will multiply each value of X by its corresponding probability and sum all of these products: \(E(X) = (1)(\dfrac{6}{36}) + (2)(\dfrac{5}{36}) + (3)(\dfrac{4}{36}) + (4)(\dfrac{3}{36}) + (5)(\dfrac{2}{36}) + (6)(\dfrac{1}{36})\) \(E(X) = (\dfrac{6}{36}) + (\dfrac{10}{36}) + (\dfrac{12}{36}) + (\dfrac{12}{36}) + (\dfrac{10}{36}) + (\dfrac{6}{36})\) \(E(X) = \dfrac{56}{36}\) Therefore, the expected value of the random variable X, which is the lower number when two dice are rolled, is \(\dfrac{56}{36}\).