Unlikely event

Calculus Level 4

Once I read a magazine article about probability where the author stated there was an unlikely event. The probability was given as $\large \left(1-10^{-21}\right)^{10^{25}}.$ This probability is pretty small but hard to get a handle on. It is, however, quite close to $10^{-n}$ , where $n$ is an integer.

Give this integer.

Note: It is easy to find websites that can evaluate this expression. Please don't. The intention is that it be solved with nothing but a scientific calculator and BRILLIANT reasoning.

The answer is 4343.

2 solutions

Jeremy Galvagni
Apr 8, 2018

$(1-\frac{1}{x})^{x}$ tends to $\frac{1}{e}$ as $x$ increases. This $x$ is very large. We can rewrite the expression as

$((1-10^{-21})^{10^{21}})^{10^{4}}$

and then the excellent approximation

$(\frac{1}{e})^{10000}=e^{-10000}$

My calculator can't handle this so I'll use a base 10 logarithm

$\log{e^{-10000}} = -10000\log{e} \approx -4342.944819$

So the probability is very close to $10^{-4343}$ and so $\boxed{n=4343}$

Chew-Seong Cheong
Apr 9, 2018

Similar solution with @Jeremy Galvagni 's

$\left(1-10^{-21}\right)^{10^{25}} = \left(1-\frac 1{\color{#3D99F6}10^{21}}\right)^{10^4\cdot \color{#3D99F6}10^{21}} \approx \displaystyle \lim_{x \to \infty} \left(1-\frac 1x\right)^{10^4x} = e^{-10^4} = 10^{-10^4\log_{10} e} \approx 10^{-4343}$

Therefore, $n = \boxed{4343}$ .

Unlikely event

The answer is 4343.

2 solutions

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