Q:
(a) Say that P(E) = 0.8 and P(F) = 0.6 Show that P(E ∩ F) ≥ 0.4.
(b) Show that if E and F are independent then so are E’ and F’.
A: Probability is my favorite.
Let me define some notation first so we are all on the same page:
P(E) is the probability of set E.
E’ is the set “E compliment” (everything that is not in E)
E U F = “E union F” which means either E or F (or both)
E ∩ F = “E intersection F” which means E and F (must be both)
OK. We start:
(a) Picture (or draw) a Venn Diagram with two circles. One circle is E and the other is F… they overlap a little.
What is the smallest possible overlap that is available? Well:
P(E) + P(F) = 1.4
Hmmm… That means there must be some overlap and that overlap must be at least .4… Why must it be at least .4? Because if E and F took up the whole sample space, then P(E) + P(F) = 1. Remember the probability of the whole space is always 1. Therefore, P(E ∩ F) must be at least .4, but could be more!
(b) If E and F are independent, this means their outcomes do not affect each other. Do not confuse independent with mutually exclusive. Let me go on a quick tangent: two things that are mutually exclusive cannot both happen at the same time (this look like two disconnected/non-overlapping circles on a Venn Diagram).
Example: “Raining” and “Not Raining” are mutually exclusive. They are also dependent: if it is “Raining” it cannot be “Not raining”. Moral of the story: mutually exclusive events are always dependent.
OK. Now we know that 2 independent events must overlap. They are independent because they do not affect each other. If two events are independent, then we know the following fact by definition:
P(E ∩ F) = P(E)*P(F)
So, we want to prove that knowing the above will give us that P(E’ ∩ F’) = P(E’)*P(F’)… Knowing this will tell us that E’ and F’ are independent as well.
Start with the following which we know is a truths (look these up in your book, it is true, I promise):
(1) P(E’ U F’) = P(E’) + P(F’) – P(E’ ∩ F’)
And:
(2) P(E’ U F’) = 1 – P(E ∩ F)
In words, the above equation says: The probability of everything not in E or not in F equals 1 minus everything in E and F.
Substitute equation (2) into (1) to get:
1 – P(E ∩ F) = P(E’) + P(F’) – P(E’ ∩ F’)
Remember from before? P(E ∩ F) = P(E)P(F)… So, sub that in:
1 – P(E)P(F) = P(E’) + P(F’) – P(E’ ∩ F’)
Again, by definition: P(E) = 1 – P(E’) and P(F) = 1 – P(F’)… Substitute that in like so:
1 – (1 – P(E’))(1 – P(F’)) = P(E’) + P(F’) – P(E’ ∩ F’)
Multiply it out with some fancy algebra:
1 – (1 – P(F’) – P(E’) + P(E’)P(F’)) = P(E’) + P(F’) – P(E’ ∩ F’)
1 – 1 + P(F’) + P(E’) – P(E’)P(F’) = P(E’) + P(F’) – P(E’ ∩ F’)
P(F’) + P(E’) – P(E’)P(F’) = P(E’) + P(F’) – P(E’ ∩ F’)
Subtract P(F’) and P(E’) from both sides:
– P(E’)P(F’) = – P(E’ ∩ F’)
P(E’)P(F’) = P(E’ ∩ F’)
Therefore, by definition, E’ and F’ are independent.