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
.
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.