SOLUTION: Two dice are cast: E1 is the event that a6 appears on at least one die. E2 is the event that a 5 appears on exactly one die and E3 is the event that same number appears on both dic

Question 1209322: Two dice are cast: E1 is the event that a6 appears on at least one die. E2 is the event that a 5 appears on exactly one die and E3 is the event that same number appears on both dice.
(i) Are E1 and E2 independent?
(ii) Are E2 and E3 independent?
(iii) Are E3 and E1 independent?

Answer by Edwin McCravy(20064) (Show Source): You can put this solution on YOUR website!

There are 36 possible rolls of the two dice.

E1 = {16,26,36,46,56,66,65,64,63,62,61}

E2 = {15,25,35,45,54,53,52,51}

E3 = {11,22,33,44,55,66}}

To find if they are independent or not, we check to see if the probability of
one of them occurring equals or doesn't equal the probability when the other is
given.

P(E1) = 11/36
P(E1 | E2) = P(E1 & E2)/P(E3) = 0/(6/36) = 0 for E1 and E2 have no elements in common.
They are not equal, so they are not independent
In fact, they are mutually exclusive (they mutually exclude each other)

P(E2) = 8/36 = 2/9
P(E2 | E3) = P(E2 & E3)/P(E3) = 0/(6/36) = 0 for E2 and E3 have no elements in common.
They are not equal, so they are not independent
In fact, they are mutually exclusive (they mutually exclude each other)

P(E3) = 6/36 = 1/6
P(E3 | E1) = P(E3 & E1)/P(E1) = P(6,6)/P(E1) = (1/36)/(11/36) = 1/11.
They are not equal, so they are not independent
But they are not mutually exclusive, either, because neither is excluded in the
case of double 6's.

So the answer to all three is "no".

----------------------------------------

You could also do the problem this way:

P(E2) = 6/36 = 1/6
P(E2 | E1) = P(E2 & E1)/P(E1) = 0/(11/36) for E2 and E1 have no elements in common.
They are not equal, so they are not independent
In fact, they are mutually exclusive (they mutually exclude each other)

P(E3) = 8/36 = 2/9
P(E3 | E2) = P(E2 & E3)/P(E3) = 0/(1/6) = 0 for E2 and E3 have no elements in common.
They are not equal, so they are not independent
In fact, they are mutually exclusive (they mutually exclude each other)

P(E1) = 11/36 
P(E1 | E3) = P(E1 & E3)/P(E3) = P(6,6)/(8/36) = (1/36)/(8/36) = 1/8.
They are not equal, so they are not independent
But they are not mutually exclusive, either, because neither is excluded in the
case of double 6's.

So the answer to all three is "no".

Edwin

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