I'm struggling with Bayes' theory. Please consider the following:(adsbygoogle = window.adsbygoogle || []).push({});

Example 1:

Submarine sinks if one missile hits it. Two ships aim at submarine and fire one missile each. Ship 1 shoots missile X1, ship 2 shoots missile X2.

P(X1 hitting = 0.8)

P(X2 hitting = 0.5)

P(X1 and X2 hitting | submarine seen sinking) = P(submarine seen sinking | X1 and X2 hitting)*P(X1 and X2 hitting)/P(submarine seen sinking)

P(X1 and X2 hitting | submarine seen sinking) = 1*(.4)/(1-.5*.2) = .4/.9 = .444444...

Example 2:

Same as example one except:

P(X1 hitting = 0.5)

P(X2 hitting = 0.5)

Thus,

P(X1 and X2 hitting | submarine seen sinking) = P(submarine seen sinking | X1 and X2 hitting)*P(X1 and X2 hitting)/P(submarine seen sinking)

P(X1 and X2 hitting | submarine seen sinking) = 1*(.25)/(.75) = .3333....

However, some would argue that in example two, P(X1 and X2 hitting | submarine seen sinking) = 0.5 - since you could say we know for sure that 1 hit, thus the probability that the other one hit is still 0.5 (independent events). Who is correct?

The reason I showed the first example, is because I don't see how you could come up with an answer without Bayes' in the first example - you don't know which one hit???

Let me show you another example which I posted in an earlier thread (https://www.physicsforums.com/showthread.php?t=607946):

Example 3:

2 fair dice. What is the probability of both showing six if I have observed

at least one six.

Thus, the way I did it was:

P(2 6's | observing at least 1 6) = P(observing at least 1 6 | observing 2 6's)*P(2 6's)/P(observing at least 1 6)

P(2 6's | observing at least 1 6) = 1*(1/36)/(1-25/36) = 1/11

However, as the person who replied to my post pointed out, an answer of 1/6 here would appear to be the logical answer - knowing that the first dice came up 6, the probability that the second dice came up 6 is still 1/6 - independent events.

So, as you can see, I really am struggling with Bayes' theory here. Could somebody please help me out of the darkness!

Thanks.

Nick.

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# Bayes' Theory Confusion

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