If the first dice yields an even number, what is the probability that the sum of the two dices will be equal to 12?

1st Die
P(even) = 3/6 = 0.50
P(6|even) = 1/3 = 0.333

2nd Die
P(6) = 1/6 = .1667

So P(12) = 0.5*0.33*0.1667 ?

To calculate the probability that the sum of the two dice will be equal to 12 given that the first dice yields an even number, you need to multiply the probability of rolling an even number on the first die, the probability of rolling a 6 on the first die given that it is even, and the probability of rolling a 6 on the second die.

Let's break down the calculations:

P(even) = 3/6 = 0.5
This is the probability of rolling an even number on the first die. Out of the six possible outcomes (1, 2, 3, 4, 5, 6), three of them are even (2, 4, 6).

P(6|even) = 1/3 = 0.333
This is the conditional probability of rolling a 6 on the first die given that it is even. Out of the three possible even numbers (2, 4, 6), only one of them is 6.

P(6) = 1/6 = 0.1667
This is the probability of rolling a 6 on the second die. Out of the six possible outcomes (1, 2, 3, 4, 5, 6), only one of them is 6.

To calculate the probability of rolling a sum of 12 given these conditions, you multiply the three probabilities together:

P(12) = P(even) * P(6|even) * P(6)
= 0.5 * 0.333 * 0.1667
= 0.0278

So the probability of rolling a sum of 12 given that the first dice yields an even number is approximately 0.0278 or 2.78%.