To find the probability that the spinner landed on an even-numbered side given that Raj's score is 12, we use conditional probability:

Let \(E\) be the event that the spinner lands on an even-numbered side. We need to find \(P(E \mid 12)\).

First, calculate \(P(E \text{ and } 12)\):

The probability of landing on an even-numbered side is \(\frac{2}{5}\) (since there are 2 even numbers: 2 and 4).

The possible dice outcomes that multiply to 12 are: (2,6), (6,2), (3,4), (4,3). Each pair has a probability of \(\frac{1}{36}\) (since there are 36 possible outcomes when rolling two dice).

Thus, \(P(E \text{ and } 12) = \frac{2}{5} \times \frac{4}{36} = \frac{8}{180} = \frac{2}{45}\).

Next, calculate \(P(12)\):

For an odd-numbered side, the possible dice outcomes that add to 12 are: (6,6). The probability of this is \(\frac{1}{36}\).

The probability of landing on an odd-numbered side is \(\frac{3}{5}\) (since there are 3 odd numbers: 1, 3, 5).

Thus, \(P(12) = \frac{3}{5} \times \frac{1}{36} + \frac{8}{180} = \frac{11}{180}\).

Finally, use the conditional probability formula:

\(P(E \mid 12) = \frac{P(E \text{ and } 12)}{P(12)} = \frac{\frac{8}{180}}{\frac{11}{180}} = \frac{8}{11}\).