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) = 0.33*0.1667 ?
If the first die = 2 or 4, the sum of the two will not = 12. The probability of a 6 = 1/6
I'm sorry but I do not follow. The question asks that (out of two dice) given the first dice is even -
In which the first die will therefore be 2,4, or 6 - where 6 is the only option.
And so what is the probability that the sum of the two dice will equal 12 given the first dice is even. So it is not simply 1/6*1/6
To find the probability that the sum of the two dice will be equal to 12 given that the first dice yields an even number, we need to find the joint probability.
First, let's calculate the probability of the first die showing an even number. Since there are 6 possible outcomes and 3 of them are even numbers (2, 4, 6), the probability of the first dice yielding an even number is 3/6 or 0.50.
Now, let's calculate the conditional probability of rolling a 6 on the second die given that the first die showed an even number. Since there are 3 even numbers and one of them is a 6, the probability of rolling a 6 on the second die given that the first die is even is 1/3 or 0.333.
Lastly, let's calculate the probability of rolling a 12 as the sum of the two dice. The probability of rolling a 6 on the second die is 1/6 or 0.1667.
To find the joint probability, we multiply the conditional probability of rolling a 6 on the second die given that the first die showed an even number by the probability of rolling a 6 as the sum of the two dice. So the probability of the sum of the two dice being 12 given that the first die is even is:
P(12|even) = 0.333 * 0.1667 = 0.0555 or 5.55%