Question 1137820: Suppose that a single fair die is rolled one time. Let events A and B be defined as:
A = { 1, 4, 5 } and B = { 2, 4, 5, 6 }
Are A and B independent events? Show why or why not? I cant figure this problem out
Found 3 solutions by Edwin McCravy, jim_thompson5910, ikleyn:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website! The intuitive method:
Two events A and B are independent of each other if and only if the probability
of either event would not change if it were known that the other event has
occurred. That is: P(A|B) = P(A) and P(B|A) = P(B)
P(A) = P( { 1, 4, 5} ) = 3/6 = 1/2,
P(B) = P( {2, 4, 5, 6 }) = 4/6 = 2/3,
P(A & B) = P( {4} ) = 1/6
P(A|B) = P(A & B)/P(B) = (1/6)÷(2/3) = (1/6)(3/2) = 3/12 = 1/4
P(A|B) = 1/4 is not equal to P(A) = 1/2,
So since the knowledge of B's occurrence would cause the probability
of A to increase from 1/4 to 1/2, they are not independent.
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There is another way but it is not intuitive:
Two events A and B are independent if P(A & B) = P(A) × P(B)
P(A) = P( { 1, 4, 5} ) = 3/6 = 1/2,
P(B) = P( {2, 4, 5, 6 }) = 4/6 = 2/3,
P(A & B) = P( {4} ) = 1/6
P(A) × P(B) = (1/2)(2/3) = 1/3
So they are not independent by the non-intuitive method.
Edwin
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
I'm assuming the set notation refers to the outcome of rolling the die
A = {1,4,5} = rolling a 1, 4 or 5
B = {2,4,5,6} = rolling a 2, 4, 5, or 6
There are 3 ways to roll an outcome in event A, out of 6 total, so
P(A) = 3/6 = 1/2
There are 4 items in set B, so
P(B) = 4/6 = 2/3
If A and B were independent, then this equation would be true
P(A and B) = P(A)*P(B)
P(A and B) = (1/2)*(2/3)
P(A and B) = 1/3
We'll keep this value in mind
Let
C = the event in which both A and B happen at the same time
In other words, event C is rolling either a 4 or 5 since these values are found in both set A = {1,4,5} and set B = {2,4,5,6}. We can say that set C is the intersection of A and B
C = {4,5}
We have 2 ways to get a value in set C, out of 6 ways to roll a die
P(C) = 2/6
P(C) = 1/3
we get 1/3 just like with the previous computation
We see that P(A and B) = P(C) = P(A)*P(B) is true, so therefore we have shown A and B are independent events.
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
In his solution, Edwin made en error, which led him to wrong conclusion.
In my post, I brought a correct solution.
Two events A and B are independent of each other if and only if P(A)*P(B) = P(A & B).
P(A) = P( { 1, 4, 5} ) = 3/6 = 1/2,
P(B) = P( {2, 4, 5, 6 }) = 4/6 = 2/3,
P(A & B) = P( {4, 5} ) = 2/6 = 1/3. <<<---=== It is where Edwin made his mistake.
Now, P(A)*P(B) =  =  and P(A & B) = .
Thus P(A)*P(B) = P(A & B), so the events A and B are independent.
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