Probability formula for life

If there are 300 billion stars in an average galaxy and 100 billion
galaxies in the observable universe, the total number of stars is 3 X
10^22. Suppose we wanted to know to a certainty of 99.9999% that life
has formed at *least* once on some planet other than earth at one of
these stars. What would be the minimum average probability of life
forming at any of these stars? We can find a simple formula to answer
this question.

Let p_a = the average probability of a star supporting life

P = the probability of at least one other star in the universe
with life

Q = 1-P = the probability of no other stars than
the sun with life

We want to find p_a given P = 0.999999 (a certainty of 99.9999%)

The probability of a star *not* supporting life is

1 - p_a

So the probability of no other stars in the universe supporting life is

(1). (1 - p_a) ^ (N-1) = Q where N = 3 X 10^22

The -1 results because we're excluding our own sun, but N is so large
that N-1 is equal to N for practical purposes.

Taking the natural logarithm of both sides of equation (1), substituting
N for N-1, and dividing through by N we get

(2). ln(1 - p_a) = ln Q / N

Using the power series expansion for ln,

(3). ln(1 - p_a) = -p_a - p_a^2 / 2 - p_a^3 / 3 - ....

Since p_a is small, we can neglect all terms but the -p_a term.

So the final equation is

(4). p_a = -ln Q / N

Now, given P = 0.999999, Q = 1-P = 10^-6

p_a = - ln(10^-6) / (3 X 10^22) = 4.61 X 10^-22

This means that if the average probability that a star in the universe
contains a planet with life is no less than 4.61 X 10^-22, we can be
99.9999% certain that the universe contains *at least* one star (other
than our sun) with life.

Those are pretty good betting odds.

Note that the above analysis ignores *when* life develops at a star.
The analysis is not constrained to extant life. If life developed only
once at some other star 200 million years ago, then disappeared, that is
an example of a successful life event.


--dkomo@xxxxxxxx






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