math: calculating probability

[JEL]

You know when you cannot see the forest for all the trees in it? I'm having such a situation with a math-question at the moment I'm hoping someone can see it more clearly:

We have 1 engine with a 50 percent chance of failure during a run.

If we mount 10 such engines on 1 vehicle, the chance of all engines failing would be 50 divided by 10... is that correct? IE 5 percent chance of total engine failure.

It's 50% chance per engine, and we have 10 of them.
'Single instance chance' divided by combined number of 'single instance items' equals 'combined chance'.

Is that how simple probability is calculated?

To take it a bit further:

We have 1 vehicle we need to protect from entry-heat.

We have 2 options of heat-shield protection:

Option #1: 1 single heat-shield giving 100% protection. If that shield is lost, the vehicle is lost. The shield has a 50% chance of failure.

Option #2: 4 heat-shields each giving 25% protection. If 1 shield is lost, the vehicle is still safe. However, if 2 or more are lost, the vehicle is lost. Each single shield has a 50% chance of failure.

Therefore, my math tells me (if the trees are not in the way that is ):

Option #1 gives the vehicle a 50% chance of failure during entry.

Option #2 gives the vehicle a 37.5% chance of failure during entry.

Does that seem correct?

(This is not part of an exam or anything )

 

[tblaxland]

If we mount 10 such engines on 1 vehicle, the chance of all engines failing would be 50 divided by 10... is that correct? IE 5 percent chance of total engine failure.

No, it is 0.5^10 = 0.098%.

See here: http://en.wikipedia.org/wiki/Joint_distribution#Joint_distribution_for_independent_variables

We have 2 options of heat-shield protection:

Option #1: 1 single heat-shield giving 100% protection. If that shield is lost, the vehicle is lost. The shield has a 50% chance of failure.

Option #2: 4 heat-shields each giving 25% protection. If 1 shield is lost, the vehicle is still safe. However, if 2 or more are lost, the vehicle is lost. Each single shield has a 50% chance of failure.

Therefore, my math tells me (if the trees are not in the way that is ):

Option #1 gives the vehicle a 50% chance of failure during entry.

Option #2 gives the vehicle a 37.5% chance of failure during entry.

Does that seem correct?

Option #1: 50% probability of loss of vehicle (LOV).

Option #2: Depends on how the heat shields are constructed, see below.

Option #2a: If they are protecting different parts of the vehicle. In that case, the probability of any two heat shields failing is: The probability of heat shield A failing, times the probability of heat shield B failing, times the probability of heat shield C not failing, times the probability of heat shield D not failing, times 6 (ie, the number of permutations that give two failed heat shields and two not failed heat shields). So 0.5*0.5*(1-0.5)*(1-0.5)*6 = 37.5% probability of LOV

Option #2b: If they are layered. In that case the probability is: the probability the first layer failing times the probability of the second layer failing. So 0.5*0.5 = 25% probability of LOV.

 

[Hielor]

Option #2: Depends on how the heat shields are constructed, see below.

Option #2a: If they are protecting different parts of the vehicle. In that case, the probability of any two heat shields failing is: The probability of heat shield A failing, times the probability of heat shield B failing, times the probability of heat shield C not failing, times the probability of heat shield D not failing, times 6 (ie, the number of permutations that give two failed heat shields and two not failed heat shields). So 0.5*0.5*(1-0.5)*(1-0.5)*6 = 37.5% probability of LOV

Option #2b: If they are layered. In that case the probability is: the probability the first layer failing times the probability of the second layer failing. So 0.5*0.5 = 25% probability of LOV.

I think option 2a might have somewhat missed the point--if you lose 3 heat shields you will still lose the vehicle, so you don't need the C and D not-failing part. More generally: You calculated the chances of any two heat shields failing, but you actually need the chances of two or more heat shields failing.

 

[tblaxland]

I think option 2a might have somewhat missed the point--if you lose 3 heat shields you will still lose the vehicle, so you don't need the C and D not-failing part. More generally: You calculated the chances of any two heat shields failing, but you actually need the chances of two or more heat shields failing.

Maths fail :facepalm:

Yes, you are right. For 2a you want the sum of the probability of exactly 2 failures (37.5%), plus the probability of exactly 3 failures (25%) plus the probability of exactly 4 failures (0.0625%). Therefore probability of LOV is 68.75%.

 

[agentgonzo]

Maths fail :facepalm:

Yes, you are right. For 2a you want the sum of the probability of exactly 2 failures (37.5%), plus the probability of exactly 3 failures (25%) plus the probability of exactly 4 failures (0.0625%). Therefore probability of LOV is 68.75%.

It will be slightly easier to calculate:
1 - (prob-of-exactly-0-failures) - (prob-of-exactly-1-failure)

 

[tblaxland]

OK, next probability question, I'll just sit on the side lines

 

[Urwumpe]

It will be slightly easier to calculate:
1 - (prob-of-exactly-0-failures) - (prob-of-exactly-1-failure)

And add the probability of exactly 0 and exactly 1 failure happening.

(OR in probability means: p(A) + p(B) - p(A AND B) )

In this case, p (A AND B) is of course zero, but in other calculations, when you deal with redundancy, this term becomes VERY important. If you don't, you will get to the same reliability estimates as the NASA managers about the space shuttle (who effectively calculated the chance of all parts, critical or not, failing in the same mission)

For example if you have "2 of 3" redundancy, the chance that only one subsystem out of three fails is p(A) + p(B) + p(C) - p(A AND B) - p(A AND C) - p(B AND C) - p(A AND B AND C).

And like always: Probabilities are always only as good as the model they are based on.

 

[agentgonzo]

For example if you have "2 of 3" redundancy, the chance that only one subsystem out of three fails is p(A) + p(B) + p(C) - p(A AND B) - p(A AND C) - p(B AND C) - p(A AND B AND C).

No, that's completely wrong.
If you have three independent events, the probability that ONLY one system fails is:
p(A) + p(B) + p(C) - 2p(A AND B) - 2p(A AND C) - 2p(B AND C) + 3p(A AND B AND C)

The probability that AT LEAST one system fails is:
p(A) + p(B) + p(C) - p(A AND B) - p(A AND C) - p(B AND C) + p(A AND B AND C).

It's easy to visualise if you draw Venn diagrams.

 

[Urwumpe]

No, that's completely wrong.
If you have three independent events, the probability that ONLY one system fails is:
p(A) + p(B) + p(C) - 2p(A AND B) - 2p(A AND C) - 2p(B AND C) + 3p(A AND B AND C)

The probability that AT LEAST one system fails is:
p(A) + p(B) + p(C) - p(A AND B) - p(A AND C) - p(B AND C) + p(A AND B AND C).

It's easy to visualise if you draw Venn diagrams.

Argh... correct. I had missed the coefficients in the combinations. I didn't look at the mess since I passed the exam.

 

[agentgonzo]

Argh... correct. I had missed the coefficients in the combinations. I didn't look at the mess since I passed the exam.

... and also got the sign of the final term wrong.

 

[Urwumpe]

... and also got the sign of the final term wrong.

Like I said, the coefficient (-1). For four variables the final term should be a +1 again.

 

[JEL]

@tblaxland; Ok, now I get it Multiply the probability of the first engine (0.5 since that equals 50 percent) with that of the next engine with that of the next engine with that of the next, and so on through all 10 engines: 0.5^10 = 0.09765625 percent chance that all 10 will fail at the same run when each engine has a 50 percent chance of failure per run.

Thanks for clearing that up

I can see the 2nd example is a bit more confusing to others than just myself, haha

Let me try to break it down;

Chance of 1 heat-shield failing is 50% (0.5)
Chance of 2 heat-shields failing is 25% (0.5*0.5 = 0.25)
Chance of 3 heat-shields failing is 12.5% (0.5*0.5*0.5 = 0.125)
Chance of all 4 heat-shields failing is 6.25% (0.5*0.5*0.5*0.5 = 0.0625)

Since there will be LOV at 2, 3 or 4 failed shields I add the 3 chances together: 25 + 12.5 + 6.25 = 43.75% chance of LOV in this particular configuration.

Hmm... is that really right?

For some reason I now think 75% seems more accurate: There is 25% chance of 2 shields failing together... there are 4 shields... 2 or more need to fail for LOV... 25+25+25=75%

What's wrong?

The chance of 10 engines failing together was 0.09765625% because they all had to fail together in linear order. The heat-shields don't have to fail together in linear order, it's enough if just 2 random shields fail, so shouldn't we multiply the chance of 2 failing, with the number of possible double-failures we can have from 4 shields? IE 25+25+25 since we can have 3 independent follow-up failures following when 1 shield has failed initially. There are 3 shields that can randomly be the next to fail after the first shield fail, so wouldn't that suggest that we have 3 times the chance of 25% (IE 75% chance of LOV)?

What do you think?

 

[agentgonzo]

Let me try to break it down;

Chance of 1 heat-shield failing is 50% (0.5)
Chance of 2 heat-shields failing is 25% (0.5*0.5 = 0.25)
Chance of 3 heat-shields failing is 12.5% (0.5*0.5*0.5 = 0.125)
Chance of all 4 heat-shields failing is 6.25% (0.5*0.5*0.5*0.5 = 0.0625)

Since there will be LOV at 2, 3 or 4 failed shields I add the 3 chances together: 25 + 12.5 + 6.25 = 43.75% chance of LOV in this particular configuration.

Hmm... is that really right?

No, I'm afraid it's not.

Chance of 1 heat-shield failing is 50% (0.5)

This is not correct. The chance of one specific heat-shield failing is 50%.
The chance that only one and only one heat-shield failing can be summed up as follows (naming heat-shields A-D):
Prob(A fails and B, C, D do not) + Prob(B fails and A, C, D do not) + Prob(C fails and A, B, D do not) + Prob(D fails and A, B, C do not).
As the probabilities that each heat shield fails is the same (and 50%), we get this equal to
(0.5 * 0.5*0.5*0.5) + (0.5 * 0.5*0.5*0.5) + (0.5 * 0.5*0.5*0.5) + (0.5 * 0.5*0.5*0.5)
= 4 * (0.5 * 0.5*0.5*0.5) = 0.25 = 25%.

You can do this similarly for the cases where 0, 1, 2, 3 and 4 heat-shields fail, giving:
Prob(0 fail) = 0.5 ^ 4 = 6.25%
Prob(exactly 1 fail) = 4 * 0.5 * 0.5^3 = 25%
Prob(exactly 2 fail) = 6 * 0.5^2 * 0.5^2 = 37.5%
Prob(exactly 3 fail) = 4 * 0.5^3 * 0.5 = 25%
Prob(exactly 4 fail) = 0.5 ^ 4 = 6.25%.

So Prob(at least 2 fail) = 37.5% + 25% + 6.25% = 68.75%

 

[EtherDragon]

This sounds like something for a logic table - then we can assign odds for each item.

I'll use CAPITAL letters to denote a "yes" and lower-case to denote a "no" for item failure. Each outcome has a 6.25% chance of occuring - with 16 outcomes we end up with our correct probability field. (100%)

"abcd" = no failures, no LOV, this particular condition has a 6.25%, as does each subsequent one... Now on to the chart.

abcd = safe
abcD = safe
abCd = safe
abCD = LOV
aBcd = safe
aBcD = LOV
aBCd = safe
aBCD = LOV
Abcd = safe
AbcD = LOV
AbCd = LOV
AbCD = LOV
ABcd = LOV
ABcD = LOV
ABCd = LOV
ABCD = LOV

Of these outcomes, 10 of them lead to LOV (10*6.25 = 62.5% of LOV) 6 outcomes lead to safe return (6*6.25 = 37.5%). These results are from a 50% chance of any single item failing. The field shifts a lot when you adjust the odds of these items failing.

Let's suppose that each panel has a 25% of failing. Each outcome changes greatly.

No Failures
abcd is = P(a)*P(b)*P(c)*P(d)
= 0.75*0.75*0.75*0.75
= 0.31640625 ~= 31.64%

One Failure
abcD = P(a)*P(b)*P(c)*P(D)
= 0.75*0.75*0.75*0.25
= 0.10546875 ~= 10.55%

Two Failures (simplified)
abCD = 0.75*0.75*0.25*0.25
~= 3.52%

Three Failures (simplified)
aBCD = 0.75*0.25*0.25*0.25
~= 1.17%

Four Failures (simplified)
ABCD = 0.25*0.25*0.25*0.25
~= 0.39%

With the above outcome binary chart - we have the following failure possibilities
abCD ~= 3.52%
aBcD ~= 3.52%
aBCd ~= 3.52%
aBCD ~= 1.17%
AbcD ~= 3.52%
AbCd ~= 3.52%
AbCD ~= 1.17%
ABcd ~= 3.52%
ABcD ~= 1.17%
ABCd ~= 1.17%
ABCD ~= 0.39%
FAIL ~= 26.19%

~ED

 

[agentgonzo]

abcd = safe
abcD = safe
abCd = safe
abCD = LOV
aBcd = safe
aBcD = LOV
aBCd = LOV
aBCD = LOV
Abcd = safe
AbcD = LOV
AbCd = LOV
AbCD = LOV
ABcd = LOV
ABcD = LOV
ABCd = LOV
ABCD = LOV

You made a mistake on that one. aBCd is LOV, not safe, making it 11 LOVs and 5 safes. Thus giving you the 68.75% chance of fail that I calculated above.

 

[EtherDragon]

You made a mistake on that one. aBCd is LOV, not safe, making it 11 LOVs and 5 safes. Thus giving you the 68.75% chance of fail that I calculated above.

Thanks for the correction, great catch!

~ED

 

[JEL]

Thanks a lot guys It's easy to go wrong in this.