Question

An urn contains four dice. Two have faces numbered 1,2,3,4,5, and 6; one has faces numbered 2,2,4,4, 6 and 6; and one has all six faces numbered 6. One of the dice is randomly selected from the urn and rolled. The same die rolled a second time. Calculate the probability that a 6 is rolled both times ( hint Law of total probability) Group of answer choices

Answer 1

21/72 = 0.2917 = 29.17% probability that a 6 is rolled both times

Explanation:

Two have faces numbered 1,2,3,4,5, and 6

So for these two, each with 1/4 probability of being chosen, the probability of rolling two six is given by:

(1/6)^2 = 1/36

So

`p_A = 2 * (1)/(4) * (1)/(36) = (1)/(72)`

One has faces numbered 2,2,4,4, 6 and 6;

The probability of rolling 2 faces six with this dice is:

(2/6)^2 = 4/36

This dice has 1/4 probability of being chosen. So

`p_B = (1)/(4) * (4)/(36) = (1)/(36)`

One has all six faces numbered 6.

The probabilityu of rolling two six is given by:

(6/6)^2 = 1^2 = 1

This dice has 1/4 probability of being chosen. So

`p_C = (1)/(4) * 1 = (1)/(4)`

Calculate the probability that a 6 is rolled both times

`p = p_A + p_B + p_C = (1)/(72) + (1)/(36) + (1)/(4) = (1 + 2 + 18)/(72) = (21)/(72)`

21/72 = 0.2917 = 29.17% probability that a 6 is rolled both times