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

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.

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?

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:

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