Which person has a higher probability of success?

Question

A ‘success’ is defined as picking an ‘A’ such that the polynomial $x^2 − Ax + 1$ has at least one real root. Mr. X is picking A from uniform distribution over [0, 5]. Mr. Y is picking A ∈ [0, 5] with the probability distribution function $f_A(a)$ given by : $f_A(a) = \frac{2a}{25}$ ∀a ∈ [0, 5]. Which person has a higher probability of success?

My approach: For at least one real root $D \ge 0$

$A^2-4 \ge 0$

$A \ge2$

For Mr. X

$P(2 \le A\le 5) = \frac{5-2}{5-0} = \frac{3}{5}$

For Mr. Y

$P(2 \le A\le 5) = \int_2^5 \frac{2a}{25} = \frac{6a}{25}$

How do I deal with 'a' here?

Answer 1

Your computation of the integral is not correct: $$ \int_2^5 \frac{2 a}{25} ~\mathrm{d}a = \frac{2}{25} \left[ \frac{5^2}{2} - \frac{2^2}{2} \right] = \frac{21}{25} > \frac{3}{5}. $$ Hence the probability is higher for Mr. Y.

Answer 2

Note what you need to be calculating: for $f(x)$ the pdf of $X$ and assuming Lebesgue measure, $$ P(a \le X \le b) = \int_a^b f(x) \, \mathrm{d} x $$ i.e. you actually seek $$ \int_2^5 \frac{2}{25} a \, \mathrm{d}a $$ On applying the fundamental theorem of calculus, the $a$ will no longer be present.

It appears you instead thought to integrate with respect to $x$ (hence multiplying the integrand by $3$), without thought as to what the $x$ or $f$ actually represent.