Alice and Bob each choose at random a number between zero and two. We assume a uniform probability law under which the probability of an event is proportional to its area. Consider the following events:

A: The magnitude of the difference of the two numbers is greater than 1/3.
B: At least one of the numbers is greater than 1/3.
C: The two numbers are equal.
D: Alice's number is greater than 1/3.

Find the probabilities P(A), P(B), P(C) and P(D).

Answer / Step-by-step explanation:

Before we start to answer this question, it would be nice to understand some basic definition and principles:

Random: This refers to an unpredictable pattern of arrangement.  Without a specific aim or logical procedural pattern.

Uniform Probability distribution: This is a type of probability distribution in which the outcomes of distribution irrespective of the pattern or distance results in equally likely outcomes.

Uniform distribution is one of the widely used distributions although it's one of the simplest ones. The idea behind this distribution is that, unlike the other distributions, the likelihood of the any outcome in the distribution range is the same.

Now that we have a brief knowledge of the key terms within the question, hence, to start answering the question,

(a) Starting with 0 for Alice, Bob can choose any number starting from   1 /  3 . As Alice increases, Bob has to increase his starting point as well. Therefore, after Alice passes  1 /  3 , Bob can start choosing numbers from 0. This creates 2 triangles with sides  1  / 3 . The area of them gives us the probability.

Therefore, : P = 2 ( 1/3 1/3 1/3)

                        = 4/9

(b) We can find the probability of having both the numbers less than  1  / 3 . Then we subtract this probability from 1.

We have: P = 1 - 1/3 . 1/3

                     = 8/9

(c) This is basically impossible as the area of a dot is negligible.

(d) It means that Alice needs to select a number between  1  / 3  and 1.

Therefore: P = 1 - 1 /3

                    = 2 / 3