91Ó°ÊÓ

Before we dive into calculations, it's helpful to understand the possible outcomes in this problem. The two dice will result in 36 possible outcomes, as each die has six faces and 6 * 6 = 36.

First, we'll find the probability of event A, which is the first dice landing on 6. Since it's a fair dice, the probability is 1/6. Now we need to find the probability of event B, which is getting a total of \(i\). The possible sums of the two dice are between 2 and 12. The probability of each sum is given by the number of ways to achieve that sum divided by the total possible outcomes, which is 36.

Now we need to find the probability of A and B occurring together - that is, the first dice landing on 6 and the total sum being \(i\). This can only happen if the second dice has a value equal to \(i-6\). The outcomes where this is true will only be for \(i \in \{7, 8, 9, 10, 11, 12\}\), since for all other values of \(i\), it is impossible for dice 1 to have a value of 6 and dice 2 to have a value of \(i-6\) (as dice can only have values between 1 and 6). So, we need to calculate the probability for those values only.

Now we have all the necessary information to calculate the conditional probability \(P(A|B)\). We can use the formula \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) for the values of \(i\) where it's possible to obtain a sum of \(i\) with the first dice being 6. Let's compute the conditional probability for all values of \(i\): \(P(A | i) = \frac{P(A \cap B)}{P(B)}\) For \(i \in \{2, 3, 4, 5, 6\}\), \(P(A | i) = 0\), since it is impossible to have the first dice land on 6 and have a sum of \(i\) for these values. For \(i \in \{7, 8, 9, 10, 11, 12\}\), calculate \(P(A | i)\) as follows: \(P(A | 7) = \frac{\frac{1}{36}}{\frac{6}{36}} = \frac{1}{6}\) \(P(A | 8) = \frac{\frac{1}{36}}{\frac{5}{36}} = \frac{1}{5}\) \(P(A | 9) = \frac{\frac{1}{36}}{\frac{4}{36}} = \frac{1}{4}\) \(P(A | 10) = \frac{\frac{1}{36}}{\frac{3}{36}} = \frac{1}{3}\) \(P(A | 11) = \frac{\frac{1}{36}}{\frac{2}{36}} = \frac{1}{2}\) \(P(A | 12) = \frac{\frac{1}{36}}{\frac{1}{36}} = 1\) Finally, we have the following conditional probabilities for each value of \(i\): - For \(i \in \{2, 3, 4, 5, 6\}\), \(P(A | i) = 0\) - \(P(A | 7) = \frac{1}{6}\) - \(P(A | 8) = \frac{1}{5}\) - \(P(A | 9) = \frac{1}{4}\) - \(P(A | 10) = \frac{1}{3}\) - \(P(A | 11) = \frac{1}{2}\) - \(P(A | 12) = 1\)