Life in Hell

Probability Level 1

There are 3 possible types of torment in Hell, which we will call A. B and C. More than one torment may be applied on a given day, or none at all if the denizens are lucky.

On any given day, there is a 2/3 chance that torment A will be applied.

Independent of this, there is a 2/3 chance that on that day torment B will be applied.

Independent of both of the above, there is a 2/3 chance that on that day torment C will be applied.

What are the chances that on a given day, exactly one of the three torments will be applied?

2/9 4/9 1/3 8/27

2 solutions

Denton Young
Mar 28, 2016

The chance that A is applied and B and C are not applied is 2/3 * 1/3 * 1/3 = 2/27.

The chance that A is not applied, B is applied and C is not applied is 1/3 * 2/3 * 1/3 = 2/27.

The chance that A and B are not applied and C is applied is 1/3 * 1/3 * 2/3 = 2/27.

So the chance that exactly one, unspecified, torment is applied and the other two are not = 2/27 + 2/27 + 2/27 = 6/27 = 2/9.

Moderator note:

Simple standard approach.

A possible extension: Suppose there are $n$ types of torment, and on any given day, for each of the torments independent of the others, there is a probability of $\frac{2}{n}$ of it being applied. Then the probability that exactly one of the torments is applied on a given day is

$n*\dfrac{2}{n}*\left(1 - \dfrac{2}{n}\right)^{n-1} = 2*\left(1 - \dfrac{2}{n}\right)^{n-1}$ .

Since we're dealing with Hell we can then let $n \to \infty$ , resulting in a probability of $\dfrac{2}{e^{2}} \approx 0.271$ .

Brian Charlesworth - 5 years, 2 months ago

Tom Engelsman
Dec 25, 2020

Let $P(x)$ denote the probability of $x$ torments occurring. The required probability is just:

$P(1) = 1 - P(0) - P(2) - P(3)$

where $P(0) = (1 - 2/3)^3 = \frac{1}{27}, P(2) = (2/3)^2 = \frac{4}{9}, P(3) = (2/3)^3 = \frac{8}{27}$

or $P(1) = 1 - \frac{1}{27} - \frac{4}{9} - \frac{8}{27} = 1 - \frac{21}{27} = \boxed{\frac{2}{9}}.$

Life in Hell

2 solutions

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