Julie throws a fair red dice once and a fair blue dice once. How do you calculate the probability that Julie gets a six on both the red dice and blue dice. Secondly, calculate the probability that Julie gets at least one six?

Aug 19, 2016

$P \left(\text{Two sixes}\right) = \frac{1}{36}$

$P \left(\text{At least one six}\right) = \frac{11}{36}$

Explanation:

Probability of getting a six when you roll a fair die is $\frac{1}{6}$. The multiplication rule for independent events A and B is

$P \left(A \cap B\right) = P \left(A\right) \cdot P \left(B\right)$

For the first case, event A is getting a six on the red die and event B is getting a six on the blue die.

$P \left(A \cap B\right) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$

For the second case, we first want to consider the probability of getting no sixes.

The probability of a single die not rolling a six is obviously $\frac{5}{6}$ so using multiplication rule:

$P \left(A \cap B\right) = \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{36}$

We know that if we add up the probabilities of all the possible outcomes we will get 1, so

$P \left(\text{At least one six") = 1 - P("No sixes}\right)$

$P \left(\text{At least one six}\right) = 1 - \frac{25}{36} = \frac{11}{36}$