Question

You have two events A and B such that P(A|B) = 0.4 and P(A) = 0.4. Based on this information, what can you conclude?

Answer 1

since knowing B has occurred does not change the probability of A occurring, those events are "independent"

Answer 2

in fact that is the definition of independence

Answer 3

so Event A is not dependent on Event B. @satellite73

Answer 4

event A alone
P(A) = 0.4
the probability of event A occurring is 0.4 (ie 40%)


now add in event B
we are given event B happens because P(A|B) = 0.4, but this probability is identical to P(A)
since P(A) = P(A|B), this means B does not change A

so yes, A is not dependent on B. If it were, then B would change A.

Answer 5

Thank you peeps.

Answer 6

hmm...
P(A|B) = conditional probability of A given B
=P(A\(\cap\)B) / P(B)

Answer 7

@aleale97
Can you confirm that the question says P(A|B) = 0.4 (conditional proability) and not \(P(A\cap B)=0.4\) (intersection)?

Since we have
\(P(A\cup B) = P(A)+P(B)-P(A\cap B)\)
we get
\(P(A\cap B) = P(A)+P(B)-P(A\cup B)\)
Substituting values into definition of conditional probability, P(A)=0.4, P(A|B)=0.4
\(P(A|B) = \dfrac{P(A)+P(B)-P(A\cup B)}{P(B)}\)
\( 0.4 = \dfrac{0.4+P(B)-P(A\cup B)}{P(B)}\)
Solving for P(B)
\(P(B)=\dfrac{P(A\cup B)-0.4}{0.6}\)
I am short of conclusions though.
All I can say is that B \(could\) be the sample space \(\Omega\), where P(B)=1, but this is one of the possible scenarios, and does not have to be the case.