Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 2 / 10F (Probability)

The cumulative distribution function is $F(x)=P(X\le x)$. For the differentiable distribution in the question, its probability density function is $f(x)=F^{\prime }(x)$, so probabilities over intervals are obtained by integrating $f$. The survival probability is $P(X>x)=1-F(x)$. Differentiating immediately givesThe strict inequality is consistent with the definition using $X\le x$. This relation connects the probability density function with the decline of the survival probability.

The independent uniform distributions give a constant joint probability density $1$ on the unit square. For $0<x<1$, integrate over $x<u<v^2$:For the random quadratic with uniform coefficients, real roots require its discriminant to be nonnegative, equivalently $U\le V^2$. The boundary has probability zero, and the area under this parabola isOn that event the roots are $-V\pm \sqrt{V^2-U}$, both nonpositive. Bounding both absolute values by one therefore amounts to bounding the more negative root:Since $0\le V\le 1$, squaring the inequality is legitimate and gives $U\ge 2V-1$. The admissible unit-square region is $\max (0,2v-1)\le u\le v^2$. Its probability isDividing by the real-root probability gives the requested conditional probability: