Question 137573: I need help on this question.
An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession and that the face values of the two rolls are added together. This 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 7 .
Event B : The sum is an odd number.
Write your answers as exact fractions for
P(A)=
P(B)=
Answer by Fombitz(32388)   (Show Source):
You can put this solution on YOUR website! Event A : The sum of two dice
First look at all of the possible outcomes.
First number rolled, second number rolled, then the sum.
1 1=2
1 2=3
1 3=4
1 4=5
1 5=6
1 6=7
2 1=3
2 2=4
2 3=5
.
.
.
6 1=7
6 2=8
6 3=9
6 4=10
6 5=11
6 6=12
All sums from 2 to 12 are possible out of 36 possible outcomes, with the following probabilities.
2=1/36
3=2/36
4=3/36
5=4/36
6=5/36
7=6/36
8=5/36
9=4/36
10=3/36
11=2/36
12=1/36
For the probability of the sum being greater than 7, you would add the probabilities of getting 8, 9, 10, 11, and 12.
P(A)=(5+4+3+2+1)/36=15/36 or 5/18.
Event B : The sum is an odd number.
Add up all of the probabilities for the sum to be 3, 5, 7, 9, and 11.
P(B)=(2+4+6+4+2)/36=18/36 or 1/2.
Since half of the sums are even and half of the sums are odd, you could have come to the same conclusion.
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