# What is the probability that the total of two dice will be greater than 8, given that the first dice is a 6?

*print*Print*list*Cite

The probability of total of two dice is greater than 8. and we know that the first die is 6

then the options are: (6,3) (6,4) (6,5) (6,6)

Then, we already have the first die 6, then the options for the second die would be 3,4,5, or 6

Then the probability is 4/6 =2/3

OR:

We could use the conditional probability formula:

P(E2/E1) = P(E2 and E1)/p(E1)... (1)

Let E1= first die is 6 ==> p(E1)=1/6

Let E2= total of two dice is > 8

Then E1 and E2 will be given by (6, 3) (6, 4) (6, 5) (6, 6).

There are 36 possible outcomes when we throw 2 dice.

So,

P(E1 and E2) = 4/36= 1/9

Now substitute is (1):

P(E2/E1)= (1/9)/(1/6) = 2/3

There will be a 2/3 chance, or .66667 chance, assuming that both dice have six spots on them.

If both dice have 6 spots, then 2 out of the 6 possible outcomes for the second die would give a number smaller than 8. 4 of the 6 possible outcomes would give a number larger than 8. That is because it will be a number larger than 8 if the die comes up as a 3, 4, 5 or 6.

If the second die has some other number of spots, there will be different answers, but I am assuming a reguler, 6-sided die.

The two dice could show up in 6*6 = 36 ways.

Getting more than or but not equal to 8 is in the following ways :

First die:

6 in first and 3 to 6 in 2nd : = 4ways

5 in 1st and 4 to 6 in 2nd : 3 ys:

4 in 1st and 5,or 6 in 2nd : 2 ways

3 in first and 6 in 2nd : 1 way:

So the number of possible ways of getting greater than 8 is 10 ways.

Therefore the required probability = 10/36 = 5/18