91Ó°ÊÓ

The problem is to find the probability of at least one six being rolled when two dice are rolled. The total number of outcomes when two dice are rolled is 36, as each die has six faces, so there are six possibilities for the first die and six possibilities for the second die. Now, it may be easier to compute the probability of the complementary event - the event that neither dice shows a \(6\). Then we will subtract this probability from \(1\) to find the probability of at least one six being rolled. For the second part, we are given that the two faces are different, so we have to adjust the total number of possible outcomes accordingly and again find the probability of at least one six under this condition.

If we want neither of the dice to show a six, we have \(5\) favorable outcomes for the first dice (1, 2, 3, 4, or 5) and \(5\) favorable outcomes for the second dice. The total number of favorable outcomes for this event is \(5 \times 5 = 25\). So, the probability of neither dice showing a six is given by: \[P(\text{no six}) = \frac{25}{36}\]

To find the probability of at least one six being rolled, we will subtract the probability of no six from \(1\): \[P(\text{at least one six}) = 1 - P(\text{no six})\] \[P(\text{at least one six}) = 1 - \frac{25}{36} = \frac{11}{36}\] So, the probability that at least one die shows a six when two dice are rolled is \(\frac{11}{36}\).

Now, we are given the condition that the two faces are different. In this case, there are \(6 \times 5 = 30\) possible outcomes (since for each face on the first die, there are 5 different faces on the second die).

To find the probability of at least one six when the two faces are different, we will again use the complementary event approach. In this case, neither die should show a six, but still be different. There are \(4\) favorable outcomes for the first die (1, 2, 3, or 4) and \(4\) favorable outcomes for the second die (the remaining ones different from the choice for the first die). The total number of favorable outcomes for this event is \(4 \times 4 = 16\). So, the probability of neither dice showing a six under this condition is given by: \[P(\text{no six | different faces}) = \frac{16}{30} = \frac{8}{15}\] Now, we will subtract this probability from \(1\) to find the probability of at least one six with the different faces condition: \[P(\text{at least one six | different faces}) =1 - P(\text{no six | different faces})\] \[P(\text{at least one six | different faces}) = 1 - \frac{8}{15} = \frac{7}{15}\] So, the probability of at least one die showing a six under the condition that the two faces are different is \(\frac{7}{15}\).