This is my question
The fair die is 6-sided and it is rolled twice. We define three events:
E1 = The first roll lies in the set {2, 3, 4}
E2 = The first roll lies in the set {4, 5, 6}
E3 = Sum of two rolls is 5
Are the events E1, E2, E3 independent of each other?
In order to find independance, following formula has to be satisfied:
$P (E1 \cap E2) = P(E1)P(E2)$
I already know the probabilities of both events i.e. ee1 and E2 is $\frac 3 6$.
And $P (E1 \cap E2) = \frac 1 6 $
Putting values in formula gives us :
$\frac 1 6 = \frac 3 6 *\frac 3 6 $, which is false.
So both events are not independent.
However, I am confused about finding independence between first and last event.
E3 looks like this = { (1,4), (4,1), (3,2), (2,3)}.
P(E3) = $\frac 1 6 * \frac 1 6 *4$(since there are 4 events).
But how do I find $P(E1 \cap E3)$?
Best Answer
E1={2,3,4}
E2={4,5,6}
E3={(1,4),(2,3),(3,2),(4,1)}
For intersection of E1 and E3, you need outcomes in E3 starting with 2 or 3 or 4.
So, it will consist of {(2,3),(3,2),(4,1)}, having probability $ \frac{3}{36} $
So, $P(E_1\cap E_3 $)=$ \frac{3}{36} $ = $ \frac{1}{12} $
and P(E1)*P(E3)= $ \frac{3}{6} $ * $ \frac{4}{36} $=$ \frac{1}{18} $.
So, E1 and E3 are not independent.
Similarly, go for E2 and E3.