Question 1190764
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Part (a)


A = the first die lands on 3
B = the sum of the dice is 8


P(A) = 6/36 = 1/6 because there are 6 ways to roll the second die if the first die is fixed at "3", and this is out of 36 ways to roll two dice.


P(B) = 5/36 because there are five ways to sum to 8 as shown below
2+6 = 8
3+5 = 8
4+4 = 8
5+3 = 8
6+2 = 8


P(A and B) = 1/36
We have only one way to sum to 8 such that the first die is 3


Now notice that 
P(A)*P(B) = (1/6)*(5/36) = 5/216
which does NOT match with P(A and B) = 1/36


So P(A and B) = P(A)*P(B) is a false statement


<font color=red>Therefore, the events A and B are <u>not</u> independent.</font> We consider them dependent.


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Here's another way to see why A and B aren't independent.


We found earlier that
P(A) = 1/6


If somehow we know 100% that event B happened, then we have
P(A given B) = 1/5
this is because the sample space (originally 36 items) has reduced to 5 items only


If we know event B happened, then the sample space is
2+6 = 8
3+5 = 8
4+4 = 8
5+3 = 8
6+2 = 8
i.e. the five ways to add to 8. Of those 5 ways, only one has 3 as the first item. 


P(A) = 1/6 doesn't match with P(A given B) = 1/5
We need P(A given B) = P(A) to be true if we want A and B to be independent.


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Part (b)


A = the first die lands on 3
C = the sum of the dice is 7


P(A) = 1/6 was calculated earlier.


Ways to sum to 7
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
There are 6 sums here out of 36 ways to roll two dice.
This means P(C) = 6/36 = 1/6


event A and C = first die lands on 3 AND the sum is 7
The only way for this to happen is to have the sum 3+4 = 7.
Which means,
P(A and C) = 1/36


Notice how
P(A)*P(C) = (1/6)*(1/6) = 1/36
which matches with the P(A and C)


We found that P(A and C) = P(A)*P(C) is a true equation.


<font color=red>Therefore, events A and C are independent.</font>
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