Question

If two fair dice are rolled, find the probability that the sum of the dice is 9, given that the sum is greater than 5.
The probability is

Answer 1

Answer:  2/13

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Explanation:

Refer to the dice chart below. It represents all possible ways to roll two dice and sum up the face values.

According to the table, there are...

• A = 1 copy of the sum "2"
• B = 2 copies of the sum "3"
• C = 3 copies of the sum "4"
• D = 4 copies of the sum "5"

There are A+B+C+D = 1+2+3+4 = 10 instances where we get a sum between 2 and 5. The table shows 6*6 = 36 sums total. This means there are 36-10 = 26 sums where they are between 6 and 12, aka "the sum being greater than 5".

Those 26 sums are what we focus on. Out of those 26 items, four of them are sums of "9".

The probability we want is therefore 4/26 = 2/13

Answer 2

The probability that the sum of the dice is 9, given that the sum is greater than 5 is 0.154

See attachment for the sample space of a roll of two dice.

From the attached sample space,

• The number of sum greater than 5 is: 26
• The number of sum greater than 5 that is 9 is: 6

Represent the events that a sum is greater than 5 with A, and a sum is greater than 5 is 9 with B.

So, we have:

$n(A) = 26$

$n(A\ n\ B) = 4$

The probability that the sum of the dice is 9, given that the sum is greater than 5 is then calculated using:

$Pr = \frac{n(A\ n\ B)}{n(A)}$

This gives

$Pr = \frac{4}{26}$

$Pr = 0.154$

Hence, the probability is 0.154

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