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From: briggs on 7 Mar 2005 14:20
In article <d0i7g0$lg5$1(a)news3.zwoll1.ov.home.nl>, SomeOne <celtic_246(a)nospam.hotmail.com> writes:
> Hi,
>
> I'm kinda stuck with this problem I'm having. Maybe you can help me.
>
> Let's say we have an event A. This event can occur through two
> mechanisms. The probability that event A occurs through mechanism 1 is
> P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8.

What is the probability that event A occurs through _both_ mechanism
1 and mechanism 2?

Let us make this problem a bit more physical and, perhaps, gain some
understanding...

A man stands in front of a two man firing squad.

Marksman 1 has a 90% chance of killing his target
Marksman 2 has a 80% chance of killing his target.

Both marksmen fire. What is the chance that the target dies?

Disregard the possibility that two separate non-fatal wounds are
fatal in combination.

> The question I'm struggling with is what is the probability that event A
> will occur?
>
> Now, just adding probabilities doesn't seem like a good idea because
> then we end up with a probability that is greater than 1, and that would
> be ridiculous. Also, multiplying probabilities seems silly because then
> you end up with a probability that is less than the original two.
>
> So, after long pondering this problem, I figured that the probability
> must be something like:
>
> P1+(1-P1)*P2

If the two marksmen had independent probabilities of failure then
this formula would be good.

There are other equivalent formulations. If I haven't goofed, some are:

1 - (1-P1)*(1-P2)
P2 + (1-P2)*P1
P1 + P2 - (P1*P2)

If the probabilities of failure are not independent then all of these
formulas fail. Suppose, for instance, that marksman 1 has a success
ratio of 100% at all times except at the first six minutes after the hour.

And suppose that marksman 2 has a success ratio of 100% at all times
except at the first twelve minutes after the hour.

Then the target has a 90% probability of being killed.

But if marksman 2 had a success ratio of 100% at all times except
for the next to last twelve minutes of the hour then the target is
sure to be killed.

One formula that always works is

P(A or B) = P(A) + P(B) - P(A and B)

For independent events, P(A and B) = P(A) * P(B) and this reduces
to any one of the formulas given previously.

John Briggs
From: Álvaro Begué on 7 Mar 2005 14:30
You don't have enough information to determine the probability of A. If
the mechanisms 1 and 2 are independent events, then your formulas are
correct and the probability of A is .98.

A simple way of looking at it is
P(A) = P(E1 U E2)
= P(!(!E1 & !E2)) <- de Morgan's law
= 1 - P(!E1 & !E2)
= 1 - (1-P(E1))*(1-P(E2)) <- using independence
= 1 - (1 -P(E1) - P(E2) + P(E1)*P(E2))
= P(E1) + P(E2) - P(E1*E2)

From: Trifon on 7 Mar 2005 15:23
Do both mechanisms (1 and 2) have same probability to be chosen?

P(A) = P(Mechanism1)*P1+ P(Mechanism2)*P2

If P(Mechanism1) = P(Mechanism2) = 0.5 then

P(A) = 0.5 * 0.9 + 0.5 * 0.8

From: SomeOne on 7 Mar 2005 15:32
briggs(a)encompasserve.org wrote:
> In article <d0i7g0$lg5$1(a)news3.zwoll1.ov.home.nl>, SomeOne <celtic_246(a)nospam.hotmail.com> writes:
>
>>Hi,
>>
>>I'm kinda stuck with this problem I'm having. Maybe you can help me.
>>
>>Let's say we have an event A. This event can occur through two
>>mechanisms. The probability that event A occurs through mechanism 1 is
>>P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8.
>
>
> What is the probability that event A occurs through _both_ mechanism
> 1 and mechanism 2?
>
> Let us make this problem a bit more physical and, perhaps, gain some
> understanding...
>
> A man stands in front of a two man firing squad.
>
> Marksman 1 has a 90% chance of killing his target
> Marksman 2 has a 80% chance of killing his target.
>
> Both marksmen fire. What is the chance that the target dies?
>
> Disregard the possibility that two separate non-fatal wounds are
> fatal in combination.
>
>
>>The question I'm struggling with is what is the probability that event A
>>will occur?
>>
>>Now, just adding probabilities doesn't seem like a good idea because
>>then we end up with a probability that is greater than 1, and that would
>>be ridiculous. Also, multiplying probabilities seems silly because then
>>you end up with a probability that is less than the original two.
>>
>>So, after long pondering this problem, I figured that the probability
>>must be something like:
>>
>>P1+(1-P1)*P2
>
>
> If the two marksmen had independent probabilities of failure then
> this formula would be good.
>
> There are other equivalent formulations. If I haven't goofed, some are:
>
> 1 - (1-P1)*(1-P2)
> P2 + (1-P2)*P1
> P1 + P2 - (P1*P2)
>
> If the probabilities of failure are not independent then all of these
> formulas fail. Suppose, for instance, that marksman 1 has a success
> ratio of 100% at all times except at the first six minutes after the hour.
>
> And suppose that marksman 2 has a success ratio of 100% at all times
> except at the first twelve minutes after the hour.
>
> Then the target has a 90% probability of being killed.
>
> But if marksman 2 had a success ratio of 100% at all times except
> for the next to last twelve minutes of the hour then the target is
> sure to be killed.
>
> One formula that always works is
>
> P(A or B) = P(A) + P(B) - P(A and B)
>
> For independent events, P(A and B) = P(A) * P(B) and this reduces
> to any one of the formulas given previously.
>
> John Briggs

Thanks, your physical approach is exactly what I meant. Well, that is to
say, we're not trying to shoot people here ;-) But the analogy is
remarkable. Thanks for your input it's a great help.
--
Maurice
From: Timothy Little on 7 Mar 2005 16:38
SomeOne wrote:
> Let's say we have an event A. This event can occur through two
> mechanisms. The probability that event A occurs through mechanism 1 is
> P1=0.9. The probability that event A occurs through mechanism 2 is P2=0.8.
>
> The question I'm struggling with is what is the probability that event A
> will occur?

As stated, the problem has insufficient information. It is also
ambiguous in what information it does have, because it doesn't say
whether the probabilities are conditional or unconditional.

If both probabilities are unconditional, then
P(A) = P(A and M1) + P(A and M2) - P(A and M1 and M2). We don't know
what P(A and M1 and M2) is. It could be anything between 0.7 and 1
leading to P(A) anywhere between 0.7 and 1.

If the probabilities are conditional (e.g. if mechanism M1 is present
then A has probability 0.9 of occuring),
P(A) = P(A | M1) P(M1) + P(A | M2) P(M2) - P(A | M1 and M2) P(M1 and M2),
and we only know 2 of the required 6 probabilities.

There are a few "natural" assumptions commonly made:

1) P(M1) = P(M2) = 0.5 and P(M1 and M2) = 0.
Flip a coin for which mechanism occurs.
We get P(A) = 0.85.

2) P(M1) = P(M2) = 1, which implies P(M1 and M2) = 1.
Firing squad - both mechanisms occur. In these cases
P(A | M1) = P(A and M1) and P(A | M2) = P(A and M2).

2a) P(~A | M1 and M2) = P(~A | M1) P(~A | M2)
Independent mechanisms - this is the one that leads to the formula
P1+(1-P1)*P2 for P(A) = 0.98.

2b) P(A | M1 and M2) = max(P(A | M1), P(A | M2))
One mechanism dominates the other, we get P(A) = 0.9.

3) P(M1) = P(M2) = 0.5, P(M1 and M2) = P(M1) P(M2) = 0.25.
Two independent "coin flips"

3a) ... etc


- Tim
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