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 ?
2 answers
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
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