3 real numbers are chosen independently from the interval $U[0,5]$. What is the probability that the largest number is in $[1,3]$?

Question

I'm not too familiar with continuous probability distributions so I'm unsure if my approach is correct. My first thought was to use conditional probability. I thought that the probability of this event is equivalent to solving the following:

$$Pr[\text{first number is } x \in[1,3]] \cdot Pr[\text{other two numbers are }\leq x ~|~ \text{first number is } x \in[1,3]]$$

This would give me $$\int_1^{3}\frac{1}{5}\cdot\frac{x}{5}dx=0.16$$

Does this approach make sense?

Answer 1

I think it's easier to just analyze this directly. A necessary condition is that no numbers be in $(3, 5]$. The probability that any one number is outside $(3, 5]$ (we'll call this a "failure condition") is $\frac 35$, so the probability that all three numbers are outside that range is $\frac {27}{125}$. Thus, the probability that at least one number is in the range $(3, 5]$ is $\frac{98}{125}$.

Another failure condition is that all three numbers are in $[0, 1)$. The probability that all three numbers are in $[0, 1)$ is $\frac {1}{125}$.

These failure conditions are independent. If neither failure condition is met, then the largest of your three numbers will be in $[1, 3]$. The probability of satisfying one of the failure conditions is $\frac{98}{125}+ \frac {1}{125}= \frac {99}{125}$ (we can add probabilities because it's impossible to satisfy both failure conditions simultaneously), so the probability that neither failure condition is satisfied is $\frac{26}{125}=0.208$.

Answer 2

The approach suggested in the question is flawed. It assumes that the first number chosen is the largest, and fails to take into account the possibility that the second or third number is larger than the first. There might be a way of cleaning this up, but I don't think that it is worth the effort in this case.

A more geometric approach: the choice of three numbers can be represented by a point in three-dimensional Euclidean space.

• The condition that all three numbers are in $[0,5]$ implies that the "universe" of possible outcomes is a cube with side length $5$ (with a vertex at the origin and sides along the positive axes).

• The condition that the maximum of the three numbers is less than $3$ implies that the points of interest are in a cube of side length $3$ (again, with a vertex at the origin and sides along the positive axes).

• Finally, the condition that at least one of the numbers must be larger than $1$ implies that none of the points of interest can be in a cube with side length $1$ (once again, with a vertex at the origin and sides along the positive axes).

The total space of possible outcomes is a cube with side length $5$, which has a volume of $5^3 = 125$ cubic units. A "good" outcome occurs if the randomly selected point is in the first cube, but not the second cube. This volume is $$ 3^3 - 1^3 = 26 \text{ cubic units.} $$ Thus \begin{align} &P( \text{the maximum of three numbers in $[0,5]$ is in $[1,3]$} ) \\ &\qquad\qquad= \frac{\text{volume of "good" outcomes}}{\text{total volume of possible outcomes}} \\ &\qquad\qquad= \frac{26 \text{ cubic units}}{125 \text{ cubic units}} \\ &\qquad\qquad= \frac{26}{125} = 0.208. \end{align}