Question 1035442: An ordinary (fair) dice is a cube with the numbers 1 through 6 on the sides (represented by painted spots). A dice is rolled twice in succession and that the face values of the the rolls are added together. The sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events: event A: the sum is greater than 6
Event B : the sum is not divisible by 3 and not divisible by 6.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! before you can determine the probabilities, you have to analyze all the possible outcomes.
you have 36 possible outcomes, because 6 * 6 = 36.
the probability of getting a 2 is 1+1 = 1/36
the probability of getting a 3 is 1+2 or 2+1 = 2/36
the probability of getting a 4 is 1+3 or 2+2 or 3+1 = 3/36
the probability of getting a 5 is 1+4 or 2+3 or 3+2 or 4+1 = 4/36
the probability of getting a 6 is 1+5 or 2+4 or 3+3 or 4+2 or 5+1 = 5/36
the probability of getting a 7 is 1+6 or 2+5 or 3+4 or 4+3 or 5+2 or 6+1 = 6/36
the probability of getting an 8 is 2+6 or 3+5 or 4+4 or 5+3 or 6+2 = 5/36
the probability of getting a 9 is 3+6 or 4+5 or 5+4 or 6+3 = 4/36
the probability of getting a 10 is 4+6 or 5+5 or 6+4 =3/36
the probability of getting an 11 is 5+6 or 6+5 = 2/36
the probability of getting a 12 is 6+6 = 1/36.
add up the total probabilities and you get (1+2+3+4+5+6+5+4+3+2+1)/36 = 36/36.
that's equivalent to 1, as it should be since the sum of all probabilities has to be equal to 1.
now that you know what the probabilities are, you can figure out the solution to the problem.
the problem says:
Compute the probability of each of the following events:
event A: the sum is greater than 6.
the probability that the sum is greater than 6 is equal to the probability that the sum is 7 or 8 or 9 or 10 or 11 or 12.
this is equal to (6 + 5 + 4 + 3 + 2 + 1) / 36 = 21/36.
if this is correct, then the probability that the sum is less than or equal to 6 must be equal to 36/36 - 21/36 = 15/36.
the probability that the sum is less than or equal to 6 is equal to (1 + 2 + 3 + 4 + 5) / 36 = 15/36, so it checks out ok.
Event B : the sum is not divisible by 3 and not divisible by 6.
if the number is divisible by 6, then it is also divisible by 3, so all you need to do is find the numbers that are divisible by 3.
the possible numbers are 2,3,4,5,6,7,8,9,10,11,12
of these numbers, take out the numbers that are divisible by 3 to get:
2,4,5,7,8,10,11.
the sum of the probabilities of rolling these numbers would be:
(1 + 3 + 4 + 6 + 5 + 3 + 2) / 36 = 24/36.
if this is correct, then the probability of the number being divisible by 3 or 6 would have to be 36/36 - 24/36 = 12/36.
that would be the probability of getting a 3 or 6 or 9 or 12.
the probability of that is (2 + 5 + 4 + 1) / 36 = 12/36, so it checks out ok.
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