What is the probability that the total of two dice will be greater than 8, given that the first dice is a 6?
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calendarEducator since 2009
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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.
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calendarEducator since 2008
write3,662 answers
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
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
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