**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.

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