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

pohnpei397 | Certified Educator

calendarEducator since 2009

starTop subjects are History, Literature, and Social Sciences

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.

check Approved by eNotes Editorial

## Related Questions

hala718 | Certified Educator

calendarEducator since 2008

starTop subjects are Math, Science, and Social Sciences

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

check Approved by eNotes Editorial

neela | Student

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

check Approved by eNotes Editorial