What is the probability that the total of two dice will be greater than 9 Given that the first dice is a 6?

in this question it is given that in a single throw of two dice what is the probability of this obtaining sum of number greater than the 7 if 4 appears on the first dice so from where we can solve this question is as we had two dice we have this is the one dies and thus we have the another dies so as we have this is a fore appears on the first day so we have this is for now we obtain the number is greater than 7 the sum of the number is greater than 7 for this condition this number on the second prize will be we have this is for this is 5 and this is 6a this should be greater than three hands that some will be greater than 7 now we have these are the three number and the total numbers we have total numbers we have this is one two three and thus we have 4 and 5 and 6 8 so as we have total numbers are we can say that n t here so this will be equal to 6 year now we have the number of

fabric outcomes of the four five six of these will be equal to 3 year now now we can get the probability hairs of this is we have probability of the event this will be equal to we have the number of favourable outcome / we have the number of Total outcome so this will be we have three here and we have this will be divided by 6 so we can write this as one by two here so this is we have the answer hai we have the answer is one by two thank you

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)

Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from 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.

What is the probability that the total of two dice will be greater than 9, given that the first die is a 5?

1. 2/5
2. 3/5
3. 1/3
4. 2/3

Answer (Detailed Solution Below)

Option 3 : 1/3

Concept:

Conditional Probability: The probability of an event occurring given that another event has already occurred is called a conditional probability.

The probability of occurrence of any event A when another event B in relation to A has already occurred is known as conditional probability. It is depicted by P (A|B).

\({\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

Calculation:

Let A = first die is {1, 2, 3, 4, 5, 6}

The probability of first die is a 5,

∴ P (A) = 1/6

Let B = total of two dice is greater than 9

Total outcomes when two dice are rolled = 6 × 6 = 36

Possible outcomes for A and B: (5, 5), (5, 6)

∴ P (A ∩ B) = 2/36 = 1/18

Applying the conditional probability formula we get,

\(\text{P}\left( \frac{\text{B}}{\text{A}} \right)=\frac{\text{P}\left( \text{A}\cap \text{B} \right)}{\text{P}\left( \text{A} \right)}=\frac{\left( {}^{1}\!\!\diagup\!\!{}_{18}\; \right)}{\left( {}^{1}\!\!\diagup\!\!{}_{6}\; \right)}=\frac{1}{3}\)

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