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Answer 1 · 2018-05-20T04:44:45

Event A: $\frac{5}{6}$ $5/6$
Event B: $\frac{11}{18}$ $11/18$

Explanation:

Event A: Their sum is greater than 7

We can solve through complementary events.
P(sum greater than 7) = 1- P(sum equal to 7)

What are the numbers that equal to 7?

1,6
2,5
3,4
4,3
5,2
6,1

In each case, you have $\frac{1}{36}$ $1/36$ chance of rolling those two numbers. Here's why:

Looking at 1,6

You have $\frac{1}{6}$ $1/6$ chance of rolling a 1 and you also have $\frac{1}{6}$ $1/6$ chance of rolling a 6. Since these two chances are independent events, then you multiply the numbers together. $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$ $1/6times1/6=1/36$

The $\frac{1}{36}$ $1/36$ applies to all 6 numbers. So $\frac{1}{36} \times 6 = \frac{1}{6}$ $1/36times6 = 1/6$

Therefore, P(sum equal to 7) = $\frac{1}{6}$ $1/6$

So, P(sum greater than 7) = $1 - \frac{1}{6} = \frac{5}{6}$ $1- 1/6 = 5/6$

Event B: the sum is not divisible by 4 and not divisible by 6

The largest sum you can have between two numbers is 12 ie 6+6 and the smallest number you can have is 2 ie 1+1

So numbers divisible by 4 in that range are: 4, 8, 12
and numbers divisible by 6 in that range are: 6, 12

Therefore, looking at the table below, you can see that there are 14 combinations that you cannot throw. Since each combination ie 1,3 or 6,6 and you have $\frac{1}{36}$ $1/36$ chance of throwing each combination, then P(sum is divisible by 4 and 6) is equal to $\frac{1}{36} \times 14 = \frac{7}{18}$ $1/36times14 = 7/18$

P(sum not divisible by 4 and 6) = 1-P(sum is divisible by 4 and 6)
P(sum not divisible by 4 and 6) = $1 - \frac{7}{18} = \frac{11}{18}$ $1-7/18 = 11/18$

![https://useruploads.socratic.org/Fg5eZ6JsQ8GC11vrgJ6g_sum_of_2_dice.jpg)

Answer 2 · 2018-05-20T04:55:42

$P (A) = \frac{5}{12}$ $P(A)=5/12$

$P (B) = \frac{11}{18}$ $P(B)=11/18$

Explanation:

Consider all the possible results of two dice rolls.

$1 + 1 = 2$ $1+1=2$
$1 + 2 = 3$ $1+2=3$
$\dots$ $cdots$
$2 + 1 = 3$ $2+1=3$
$2 + 2 = 4$ $2+2=4$
$\dots$ $cdots$
$6 + 5 = 11$ $6+5=11$
$6 + 6 = 12$ $6+6=12$

In total, there are $36$ $36$ combinations:

$2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12$ $2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 12$

$15$ $15$ out of these $36$ $36$ are strictly greater than 7.
$22$ $22$ out of these $36$ $36$ are not divisible by 4 or by 6.

Therefore,

$P (A) = \frac{15}{36} = \frac{5}{12}$ $P(A)=15/36=5/12$

$P (B) = \frac{22}{36} = \frac{11}{18}$ $P(B)=22/36=11/18$

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