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Evaluate: $\displaystyle\lim_{x\to0}\dfrac{x\cot(4x)}{\sin^2 x\cdot\cot^2(2x)}$
Rewrite using $\cot\theta = \cos\theta/\sin\theta$:
$\dfrac{x\cdot\cos4x/\sin4x}{\sin^2x\cdot\cos^22x/\sin^22x}$
As $x\to0$: $\sin4x\approx4x$, $\sin2x\approx2x$, $\sin x\approx x$, all cosines $\to1$:
$\approx\dfrac{x\cdot\frac{1}{4x}}{x^2\cdot\frac{1}{4x^2}} = \dfrac{\frac{1}{4}}{\frac{1}{4}} = \mathbf{1}$
A test paper has 5 questions, each with three answer choices of which only one is correct. What is the probability that a student scoring purely by guessing gets 4 or more questions right?
This is a binomial distribution with $n=5$, $p=\dfrac{1}{3}$ (correct), $q=\dfrac{2}{3}$ (wrong).
$P(X \geq 4) = P(X=4) + P(X=5)$
$P(X=4) = \binom{5}{4}\left(\dfrac{1}{3}\right)^4\left(\dfrac{2}{3}\right)^1 = 5 \cdot \dfrac{1}{81} \cdot \dfrac{2}{3} = \dfrac{10}{243}$
$P(X=5) = \binom{5}{5}\left(\dfrac{1}{3}\right)^5 = \dfrac{1}{243}$
$P(X \geq 4) = \dfrac{10}{243} + \dfrac{1}{243} = \dfrac{11}{243} = \dfrac{11}{3^5}$
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