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 ohitcome of a single trial of a random experiment. Compute the probability of each of the following events. 00 Event A: The sum is greater than 9. Event B: The sum is not divisible by 6. Write your answers as fractions.​

Answer 1

Answers:

P(A) = 1/6

P(B) = 5/6

=========================================================

Step-by-step explanation:

Let's calculate P(A).

Refer to the dice chart below.

We have

• 3 copies of "10"
• 2 copies of "11"
• 1 copy of "12"

There are 3+2+1 = 6 sums that are greater than 9.

This is out of 6*6 = 36 sums total

P(A) = 6/36 = (1*6)/(6*6) = 1/6

----------------------

Now to calculate P(B).

The multiples of 6 in the table are: 6, 12

• There are 5 copies of "6"
• There is 1 copy of "12"

There are 1+5 = 6 items that are a multiple of 6. This means we have 36-6 = 30 sums that aren't a multiple of 6

P(B) = 30/36 = (5*6)/(6*6) = 5/6

Side note: it's probably coincidence that P(A)+P(B) = 1 in this case.