Two balanced dice are rolled. Let X be the sum of the two dice.(i) Obtain the probability distribution of X (i. e. what are the possible values for X and the probability for obtaining each value?). Check that the probabilities sum to one.(ii) What is the probability for obtaining X >= 8?(iii) What is the average value of X?

i) X       2       3       4         5        6       7        8        9       10      11

P(X) 1/36   2/36  3/36  4/36  5/36  6/36  5/36  4/36  3/26  2/36

X       12

P(X)  1/36

And if we check for this case when we add the probabilities we got:

ii)

iii)Explanation:

For this case we can define the following sample space for a two balanced dice:

(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)

We have a total of 36 possible pairs

Part a

(i) Obtain the probability distribution of X (i.e. what are the possible values for X and the probability for obtaining each value?).

And the possible values ofr X on this case are : X = 2,3,4,5,6,7,8,9,10,11,12

Now we can obtain the porbability distribution for X like this:

X       2       3       4         5        6       7        8        9       10      11

P(X) 1/36   2/36  3/36  4/36  5/36  6/36  5/36  4/36  3/26  2/36

X       12

P(X)  1/36

And if we check for this case when we add the probabilities we got:

(ii) What is the probability for obtaining X >= 8?

For this case we want this probability:And if we replace we got:

(iii) What is the average value of X?

For this case we can use the definition of expected value like this and we got:

And if we replace we have:

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a.) travels faster in solids because the particles are closer together.

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