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 obtcome 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 divisible by 4. Write your answers as fractions. ​

Answer 1

Answers:

P(A) = 7/12

P(B) = 1/4

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Explanation:

Instead of having one die, let's say we have two dice. I'll make one red and the other blue.

I'll be using the dice chart shown below. The red and blue values add up to the black numbers inside the table. For instance, we have 1+1 = 2 in the upper left corner. There are 6*6 = 36 sums total.

Using that table, we can see the following:

• There are 6 copies of "7"
• There are 5 copies of "8"
• There are 4 copies of "9"
• There are 3 copies of "10"
• There are 2 copies of "11"
• There is 1 copy of "12"

In total, we have 6+5+4+3+2+1 = 21 instances where the two dice add to something larger than 6.

This is out of 36 ways to roll two dice.

Therefore P(A) = 21/36 = (3*7)/(3*12) = 7/12

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If a number is divisible by 4, then it is a multiple of 4.

The multiples of 4 found in the table are: 4, 8, 12

We have

• 3 copies of "4"
• 5 copies of "8"
• 1 copy of "12"

This gives 1+5+3 = 9 values that are a multiple of 4

P(B) = 9/36 = (1*9)/(4*9) = 1/4