SOLUTION: a pair of dice is rolled. What are the odds in favour of each of the following events occuring? a) a sum of 7 turning up b) a sum of 11 turning up c) a sum of 7 or 11 turnign up

Algebra ->  Probability-and-statistics -> SOLUTION: a pair of dice is rolled. What are the odds in favour of each of the following events occuring? a) a sum of 7 turning up b) a sum of 11 turning up c) a sum of 7 or 11 turnign up      Log On


   



Question 160937: a pair of dice is rolled. What are the odds in favour of each of the following events occuring?
a) a sum of 7 turning up
b) a sum of 11 turning up
c) a sum of 7 or 11 turnign up
d) a sum of 2 or a sum of 12 turning up
e) a sum greater than 9 turnign up

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's look at all of the possibilities
Dice 1 / Dice 2 => 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
2 / 4 => 6
2 / 5 => 7
2 / 6 => 8
3 / 1 => 4
3 / 2 => 5
3 / 3 => 6
3 / 4 => 7
3 / 5 => 8
3 / 6 => 9
4 / 1 => 5
4 / 2 => 6
4 / 3 => 7
4 / 4 => 8
4 / 5 => 9
4 / 6 => 10
5 / 1 => 6
5 / 2 => 7
5 / 3 => 8
5 / 4 => 9
5 / 5 => 10
5 / 6 => 11
6 / 1 => 7
6 / 2 => 8
6 / 3 => 9
6 / 4 => 10
6 / 5 => 11
6 / 6 => 12
There are 36 possible outcomes.
.
.
.
a) The sum of 7 shows up 6 times.
P(sum 7)=Outcomes of sum 7/Total Outcomes
P(sum 7)=6/36=1/6
.
.
.
b) The sum of 11 shows up 2 times.
P(sum 11)=2/36=1/18
.
.
.
c) They're independent so add the probabilities.
P(sum 7) or P(sum 11)=1/6+1/18=4/18=2/9
.
.
.
d)Same as c), calculate individual probabilities and add.
P(sum 2) or P(sum 12)=1/36+1/36=1/18
.
.
.
e) Let's look at all of the sums and their number of occurences.
2=>1
3=>2
4=>3
5=>4
6=>5
7=>6
8=>5
9=>4
10=>3
11=>2
12=>1
A sum greater than 9 would be 10,11, or 12.
P(sum>9)=(3+2+1)/36=6/36=1/6