Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 2 / 17

For a continuous random variable with probability density function $p$, its expected value and variance areThe mean requires an integrable first absolute moment, and a finite variance requires a finite second moment. When these hold, $\mathrm{Var}X=\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$.

Take $\sigma >0$ and substitute $u=(x-\mu )/\sigma$. The density becomes the standard Gaussian distribution density $\phi (u)=e^{-u^2/2}/\sqrt{2\pi }$. Its normalization follows from the supplied Gaussian integral after scaling by $\sqrt{2}$. Since $u\phi (u)$ is an integrable odd function, its integral is zero, and therefore $\mathbb{E}X=\mu$.

For the second moment, ${\phi }^{\prime }=-u\phi$. Integration by parts givesThe endpoint term vanishes by Gaussian decay. Consequently

The conditional probability product rule gives $P(A\mid B)P(B)=P(A\cap B)$. Applying it once more givesThe conditionals in the original expression require their conditioning events to have positive probability.

The events $B$ and $\overline{B}$ form a partition. The law of total probability therefore givesAs written, both conditionals require nonzero conditioning probabilities.

The numerator $P(B\mid A)P(A)$ equals $P(A\cap B)$, while $P(A\mid B)=P(A\cap B)/P(B)$. Thus, wherever the original quotient is defined,In particular a zero numerator and denominator cannot be cancelled; the original expression requires $P(A\mid B)>0$. This is the rearrangement of Bayes' theorem.

The two-event inclusion-exclusion principle subtracts the overlap counted twice:

This is exactly the definition of conditional probability, provided $P(A)>0$:

Under a uniform shuffle, every unordered five-card hand is equally likely, giving $\left(\genfrac{}{}{0ex}{}{52}{5}\right)$ possible hands. Choose the triple's rank in 13 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{3}\right)$ ways; then choose a different pair rank in 12 ways and its suits in $\left(\genfrac{}{}{0ex}{}{4}{2}\right)$ ways. Each full house determines these choices uniquely, so there is no further ordering factor. ThereforeThe count uses binomial coefficients because the order of cards within a hand is irrelevant.