Question

Suppose that a monkey is seated at a computer keyboard and randomly strikes the 26 letter keys and the space bar. Find the probability that its first 39 characters (including spaces) will be "to be or not to be that is the question." (Leave your answer as a formula.)

Answer 1

$|\Omega|=27^{39}\\|A|=1\\\\P(A)=\dfrac{1}{27^{39}}$

In case you want it as a fraction

$P(A)=\dfrac{1}{66555937033867822607895549241096482953017615834735226163}$

:D

Answer 2

Step-by-step answer:

There are 27 possibilities for each character typed.

The keys are striked randomly, so we assume independence between letters.

So the probability of hitting a specified character is 1/27.

Here the length of the string (excluding the final period, since it is not on the monkey's keyboard) is 39.

By the multiplication law of probability, the probability of multi-step independent events is the product of the probability of the individual events.

So probability of random typing resulting in the phrase

"to be or not to be that is the question"

= (1/27)^39

= 1/66555937033867822607895549241096482953017615834735226163