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# Life on other Planets?

Author Topic:   Life on other Planets?
cavediver
Member (Idle past 3070 days)
Posts: 4129
From: UK
Joined: 06-16-2005

 Message 16 of 160 (594354) 12-03-2010 7:29 AM Reply to: Message 1 by Bolder-dash12-02-2010 11:32 PM

...there is life on other planets?
Almost certainly (and I only include the "almost" out of due deference to tentativity)
Now, how much of this life has developed into something more complex than a basic microbial stage is another question entirely...
Basic reasoning for the above (off the top of my head):
Abiogenesis is a physical process and thus we can assign a gross probability, p, to the probability that abiogenesis occurs in a single trial.
The number of trials appropriate to p is roughly the number of stellar systems to +/- an order of magnitude or so.
Let us consider just the Observable Universe. That gives an n of around 1022. Let X be the number of abiogenetic events.
If p>10-22, then E(X)>1 and P(X=1) is very small.
If p~10-22, then E(X)~1, but still P(X>1) > P(X=1), and suggests a level of "fine-tuning" which would require further explanation.
If P<10-22, then E(X)~0 and P(X=1)>>P(X>1), but P(X=1)<<1 and thus X=1 is a highly significant event which requires further explanation.
Edited by cavediver, : No reason given.

 This message is a reply to: Message 1 by Bolder-dash, posted 12-02-2010 11:32 PM Bolder-dash has replied

 Replies to this message: Message 17 by Bolder-dash, posted 12-03-2010 7:53 AM cavediver has replied

cavediver
Member (Idle past 3070 days)
Posts: 4129
From: UK
Joined: 06-16-2005

 Message 18 of 160 (594362) 12-03-2010 8:22 AM Reply to: Message 17 by Bolder-dash12-03-2010 7:53 AM

Sorry, standard stats notation - E(X) is the Expectation of X, otherwise known as the mean in other contexts, and E(X) = n x p in this context. P() is simply the probability of the condition in the brackets.
Example with dice: X is rolling a six; n is number of rolls, p is probability of rolling a six.
So with p=1/6, n=6
E(X)=1 (i.e. we expect to get 1 six when rolling a die 6 times)
P(X=1) = 0.4 (but the probability of getting 1 six is actually only 0.4, so even though we "expect" 1 six, we are more likely to get something else!)

 This message is a reply to: Message 17 by Bolder-dash, posted 12-03-2010 7:53 AM Bolder-dash has replied

 Replies to this message: Message 19 by Bolder-dash, posted 12-03-2010 8:28 AM cavediver has replied Message 24 by Bolder-dash, posted 12-03-2010 11:01 AM cavediver has not replied

cavediver
Member (Idle past 3070 days)
Posts: 4129
From: UK
Joined: 06-16-2005

 Message 20 of 160 (594369) 12-03-2010 8:42 AM Reply to: Message 19 by Bolder-dash12-03-2010 8:28 AM

But how do you arrive at 0.4 for the probability of 1 six? Why isn't it .3? Or .45?
Count up the probabilities: six rolls, so the six could occur on any one roll.
P(six on 1st roll, and not a six on other five rolls) = 1/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6
P(six on 2nd roll, and not a six on other five rolls) = 5/6 x 1/6 x 5/6 x 5/6 x 5/6 x 5/6
P(six on 3rd roll, and not a six on other five rolls) = 5/6 x 5/6 x 1/6 x 5/6 x 5/6 x 5/6
etc...
These are mutually exclusive probabilities, so they can be added up to give
P(six on just one roll, and not a six on the other five rolls) = 0.40 (2 d.p.)
We can condense the arithmatic by using the mathematics of the Binomial Distribution:
P(X=r|n, p) = pr(1-p)(n-r)Cnr
In this case, we have
P(X=1 | 6, 1/6) = (1/6) x (5/6)5 x C61 = 0.40 (2 d.p.)
Oh, and...
$\color{white} {C^n}_r = {{n!}\over{r!(n-r)!}}$

 This message is a reply to: Message 19 by Bolder-dash, posted 12-03-2010 8:28 AM Bolder-dash has not replied

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