# What is the probability that the total of two dice will be greater than 7, given that the first die is a 5?

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We are given that the first die shows up 5 and the total of the dice is greater than 7.

Now we need to use conditional probability here. We have to determine the probability of the total being greater than 7 given that the first die is 5.

Let A denote the event that the first die is 5.

Let B denote the set that the sum of the dice is greater than 7 and the first die has a 5. Now B includes the following values of the dice, (5 , 3), (5 , 4), (5, 5) and ( 5, 6)

The possibility of getting a 5 when the first die is thrown is (1/6). So P(A) = 1/6.

The possibility of getting the 4 options that constitute the set B is 4/ 36 = 1/9

Therefore P( B|A) = P( A and B) / P(A) = (1/9) / (1/6)

= 6/ 9

= 2/3

**Therefore the required probability is 2/3.**

To find the probability that the total of two dies is greater than 7, given that the first die is 5.

Solution: Since the first event of the first die is 5, is already known, the that any number ways of the second dies comes up is 6.

The number of ways second die should show up a number that sums more than 7 with the 5 is 4. That is , second die should come up with one of the numbers 3, 4, 5, 6. So probability btaining of obtaining any of these 4 numbers is 4/6 = 2/3.

Therefore the probability of obtaining a sum of 7 in two dice given that the first dice has already shown up 5 is 2/3.

Given that first dice score is 5, the score of the second dice score must be 3 or more for the total score to be more than 7.

The second dice has can have any one of the scores 1, 2, 3, 4, 5, and 6. The probability of any one of these scores is 1/6.

It is important that since the score of the first die is specified as 5, there is no uncertainty regarding this result. Being a result that has already occurred, we take its probability as 1.

Therefore:

Probability of getting score of 3, 4, 5, or 6 = 1/6 + 1/6 + 1/6 +1/6 = 2/3

Answer:

Required probability = 2/3