If an event is impossible, its probability is what

Consider the following game: I write down a random real number between 0 and 1, and ask you to guess it. What’s the probability that you guess it correctly? The answer is zero. You might wonder: “But it’s possible for me to guess the correct answer! That means that the probability has to be more than zero!” and you would be justified in wondering, but you’d be wrong. It’s true that events that are impossible have zero probability, but the converse is not true in general. In the rest of this post, we show why the answer above was in fact zero, and why this doesn’t need to do irreparable damage to your current worldview.

Let’s begin by showing that the probability that you guess my number right is zero. Let  be the probability in question. The idea is to show that  for any positive real number . We know that , and if it’s smaller than ANY positive number, then it has to be zero! The argument is as follows. Let’s call the number I randomly picked . Imagine that the interval  is painted white. Pick any positive real number . Then there is an sub-interval of length  within the interval  containing . Imagine that this sub-interval is painted black, so now we have a black strip of length  on the original white strip, and the number I chose was in the black strip. What’s the probability that your guess lands on the black strip? It has to be , since that’s the proportion of the white strip that is covered. But in order for your guess to equal my number , it has to land in the black strip, so your probability  of guessing  can’t be larger than the probability of guessing a number on the black strip! Therefore .

You should now be convinced that this event indeed has zero probability of happening, but it’s still true. This phenomenon is because of the following geometric fact: it’s possible to have a non-empty set with zero “volume”. The term “volume” depends on the context; in the case of the point on the interval, “volume” is length. The probability of an event measured on the interval  is equal to its length, and a single point on the interval has zero length, yet it’s still a non-empty subset of the interval! Probability is basically a measure of “volume” where the entire space has “volume” equal to 1. By defining probability in this way, we can prove all kinds of neat facts using something called measure theory.

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer:   (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))