Uniform RV $X$ has $E[X] = 10$ and $P[X\geq 12] = 1/4$, calculate $P[X<9]$

Question

$X$ is a uniform random variable. It has the expected value $E[X] = 10$ and the probability $P[X\geq 12] = \frac{1}{4}$. What is the probability $P[X\leq 9]$?

Here is how I attempted it: $P[X\geq 10] = \frac{1}{2}$ because it is uniform. $P[X\geq 12]$ is $2$ away from $P[X\geq 10]$. $$(1/4)/2 = 1/8.$$ So it is $\frac{1}{8}$ between each integer. $P[X\leq 9]$ is $1$ integer away from $P[X\geq 10]$, so $$P[X\leq 9] = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}.$$

I know this is a messed up way to solve it and I don't even know if it's right. There must be a more coherent way?

Answer 1

To be serious, I guess you could assume $X$ is uniform continuously in the interval $[a,b]$ with a PDF:

$$f_X(x) = \cases{\dfrac1{b-a}&for a < x < b\\0&otherwise}$$

The expected value $E[X] = 10$, which gives one linear equation: $$\frac{a+b}{2} = 10$$

From $P[X\ge 12]$,

$$\begin{align*} \frac14 &= P[X\ge 12]\\ &= \int _{12}^b \frac{dx}{b-a}\\ &= \left[\frac{x}{b-a}\right]_{12}^b\\ &= \frac{b-12}{b-a}\\ 48 &= a + 3b \end{align*}$$

Solve for $a$, compare it with the $9$ in $P[X<9]$, and since $a = 6 < 9$,

$$\begin{align*} P[X<9] &= \int_a^9 \frac{dx}{b-a}\\ &= \left[\frac x{b-a}\right]_a^9\\ &= \frac{3}{8} \end{align*}$$

Answer 2

It looks good to me. You could spin this several ways, but I'll just branch off your approach.

Assume $X\sim \operatorname{unif}[a,b]$. Well, it has constant density, call it $h$. As you noted, we know that $$P(X> 10) = h(b-10) = \frac{1}{2}$$

(the height times the length is just the "area under the curve").

We were also given that $$P(X> 12) = h(b-12) = \frac{1}{4}.$$

Solving this system gives $h = \frac{1}{8}$ and $b = 14$. This implies $a = 6$ since $h(b-a) = 1.$

Now, $$P(X\leq 9) = \frac{1}{8}(9-6) = \frac{3}{8}.$$

Alternatively, since we know $P(X>12) = \frac{1}{4}$, then $P(X\leq 12) = \frac{3}{4}$, and by ratios $$\frac{h(12-a)}{h(10-a)} = \frac{3/4}{1/2}.$$ This immediately gives $a = 6$, and $P(X\leq 9)$ is given as above.

Answer 3

$X\sim U(a,b)$. Note that: $$\mathbb E(X)=\mu =\frac{a+b}2=10 \ \text{implies} \ \ P(10\le X\le b)=\frac12\\ \ P(X\ge 12)=\frac14 \ \ \text{implies} \ \ P(12\le X\le b)=\frac14$$ Hence, $b=14$ (because $b-10=2(b-12)$) and $a=6$ (because $a+b=20$).

Thus: $$P(X\le 9)=\frac{9-6}{14-6}=\frac38.$$