# Problem 2. Two dice are thrown and one of them turns up 5, what is the probability that the sum of the two is greater than or equal to 9.

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Two dice are thrown and one turns up a five. What is the probability that the sum is greater than or equal to nine?

We assume that both dice could be 5; that is the question could be worded at least one five turns up.

(1) There are 11 possible throws that contain a five: (5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5).

Of these, 5 have a sum of nine or greater .

The probability is 5/11

(2) We can use conditional probability to confirm:

`P(B|A)=(P(A"and"B))/(P(A))`

P(A and B) is the probability that there is a five and the sum is greater than or equal to 9. There are 5 such throws: (5,4),(5,5),(5,6),(4,5),(6,5). So the probability is 5/36.

P(A) is the probability that there is a 5 -- there are 11 such throws. So the probability is 11/36.

Thus P(B|A)=`(5/36)/(11/36)=5/11 `

** Note that if the question means exactly one five, the probability sought is 2/5 as we cannot use the roll of two fives.

**Sources:**

Two dice are thrown and **one** of them turns up 5. This could be either of the two dice that are thrown. Irrespective of which die turns up 5, the other die can turn up one of the following numbers (1,2,3,4,5,6). The sum of the two is greater than or equal to 9 in the case when the die that does not turn up 5 turns up 4, 5 or 6.

Writing the total number of cases as sets gives {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(1,5),(2,5),(3,5),(4,5),(6,5)}. When two dice are rolled there is only one case in which the outcome is (5,5) not 2.

The cases where the sum is greater than or equal to 9 are {(5,4),(5,5),(5,6),(4,5),(6,5)}. This gives the required probability as 5/11.