1. A fair die is rolled. What is the probability of getting the number 1?

    2. A.:

    Since the die is fair, all sides have equal probability. So,

    \(\Pr({\mathrm x} = 1) = \dfrac{1}{6}\),

    where \({\mathrm x}\) is the random variable equal to the number drawn.

  2. Suppose you receive the extra information that a second fair die is rolled, and the sum of the results from the first and second dice is greater than or equal to 4. What is the probability of having 1 in the first die now?

    3. A.:

    Let \({\mathrm x}\) and \({\mathrm y}\) be the random variables correspondng to the numbers drawn in the first and second dice, respectively. The problem asks for \(\Pr({\mathrm x} = 1 | {\mathrm x + y} \geq 4)\). Using the conditional probability formula, we get:

    \(\Pr({\mathrm x} = 1 | {\mathrm x + y} \geq 4) = \dfrac{\Pr({\mathrm x} = 1, {\mathrm x + y} \geq 4)}{\Pr({\mathrm x + y} \geq 4)}\).

    Now \(\Pr({\mathrm x} = 1, {\mathrm x + y} \geq 4) = 4/36\), and \(\Pr({\mathrm x + y} \geq 4) = 33/36\). Plugging these values in the above formula, we get:

    \(\Pr({\mathrm x} = 1 | {\mathrm x + y} \geq 4) = \dfrac{4/36}{33/36} = \dfrac{4}{33}\).