# When 2 dice are thrown simultaneously find the probability that the sum obtained is less than 9?

When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):

(1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |

(2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |

(3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |

(4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |

(5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |

(6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |

**the multiplication principle**: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of all possible outcomes for (a,b) is called

**the sample space**of this probability experiment.

With the sample space now
identified, formal probability theory requires that we identify the possible **events**. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration
with this example of two dice.

With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have

E={(1,4),(2,3),(3,2),(4,1)}.Note that we have listed all the ways a first die and second die add up to 5 when we look at their top faces.

Consider next the probability of E, P(E). Here we need more information. If the two dice are **fair** and **independent **, each
possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9.

In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36.

What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E.

Solution : Total number of times the two dice are thrown = 500 . <br> (i) Let `E_(1)` be the event of getting the sum 5 . Then , <br> P(getting the sum 5) <br> `= P(E_(1))` <br> `= ("number of times the sum 5 is obtained")/("total number of times the two dice are thrown")` <br> = `(56)/(500) = (14)/(125)` . <br> (ii) Let `E_(2)` be the event of getting a sum more than 9 . Then , <br> `E_(2)` = event of getting a sum 10 , 11 or 12 . <br> `therefore` P (getting a sum more than 9) <br> `= P (E_(2))` <br> `= ("number of times a sum 10 , 11 or 12 is obtained")/("total number of times the two dice are thrown")` <br> `= (53 + 29 + 28)/(500) = (110)/(500) = (11)/(50)`. <br> Let `E_(3)` be the event of getting a sum less than or equal to 6 . <br> Then , `E_(3)` = event of getting a sum 2 , 3 , 4 5 or 6 . <br> `therefore` P (getting a sum less than or equal to 6) <br> `= P(E_(3))` <br> `= ("number of times a sum , 2 , 3 , 4 , 5 or 6 is obtained")/("total number of times the two dice are thrown")` <br> `= (22 + 30 + 48 + 56 + 64)/(500) = (220)/(500) = (11)/(25)`. <br> (iv) Let `E_(4)` be the event of getting the sum between 6 and 10 . <br> Then , `E_(4)` = event of getting a sum 7 , 8 or 9 . <br> `therefore` P(getting the sum between 6 and 10) <br> `= P(E_(4))` <br> = `("number of times a sum 7 , 8 or 9 is obtained")/("total number of times the two dice are thrown")` <br> = `(70 + 64 + 26)/(500) = (160)/(500) = (8)/(25)`.