Probability problem for statisticians/mathematicians

[sasguru]

If it requires 30 space shuttle flights to build the space station, and the probability of success for each flight is 98%, what is the probability of at least one failure in the sequence of 30 flights?

This seemed to be a straightforward case of a Binomial distribution to me with:

n=30
r=1
p=0.02
q=0.98

Plugging into the Binomial formula gives me a probability of 0.333.

However, according to NASA it is 0.888.
And another text book gives 0.4545.

Can anyone explain the right answer?

[ControlG]

FWIW I go with the 0.4545.

Isn't it just

1 - (probability all successful)

= 1 - (0.98)^30

= 0.4545

[Marina]

If it requires 30 space shuttle flights to build the space station, and the probability of success for each flight is 98%, what is the probability of at least one failure in the sequence of 30 flights?

This seemed to be a straightforward case of a Binomial distribution to me with:

n=30
r=1
p=0.02
q=0.98

Plugging into the Binomial formula gives me a probability of 0.333.

However, according to NASA it is 0.888.
And another text book gives 0.4545.

Can anyone explain the right answer?

I reckon the text book is correct.

"probability of at least one failure" equals "1 - probability of no failures"

and the probability of no failures is the probability the first mission is OK AND the second mission is OK and etc for all 30 missions.

If each mission is an independent event then the ANDs are products. So the probability of as least one failure is 1 - 0.98^30 = 0.46 roughly.

P.S. I did the calculation in Google: http://www.google.com/search?hl=en&q...30&btnG=Search

Edit: It's debatable whether missions are independent events though, for this calculation, if you take into account progressive weakening due to metal fatigue and so on. Maybe 0.98 is an average chance of success, but the actual chances decline after each mission.

[sasguru]

FWIW I go with the 0.4545.

Isn't it just

1 - (probability all successful)

= 1 - (0.98)^30

= 0.4545

Aye, that's the text book answer. But I want to know why I shouldn't use the Binomial - the problem seems to fulfil all the conditions viz:

The trials must meet the following requirements:

"1. the total number of trials is fixed in advance;
2. there are just two outcomes of each trial; success and failure;
3. the outcomes of all the trials are statistically independent;
4. all the trials have the same probability of success"

[DimPrawn]

If it requires 30 space shuttle flights to build the space station, and the probability of success for each flight is 98%, what is the probability of at least one failure in the sequence of 30 flights?

This seemed to be a straightforward case of a Binomial distribution to me with:

n=30
r=1
p=0.02
q=0.98

Plugging into the Binomial formula gives me a probability of 0.333.

However, according to NASA it is 0.888.
And another text book gives 0.4545.

Can anyone explain the right answer?

I worked it out as.

1 - (0.98 raised to the power of 30) = 0.4545

[Diver]

I reckon the text book is correct.

"probability of at least one failure" equals "1 - probability of no failures"

and the probability of no failures is the probability the first mission is OK AND the second mission is OK and etc for all 30 missions.

If each mission is an independent event then the ANDs are products. So the probability of as least one failure is 1 - 0.98^30 = 0.46 roughly.

P.S. I did the calculation in Google: http://www.google.com/search?hl=en&q...30&btnG=Search

Statistically the probability of failure would increase with each trip made. the more trips the higher the probability.

[expat]

If it requires 30 space shuttle flights to build the space station, and the probability of success for each flight is 98%, what is the probability of at least one failure in the sequence of 30 flights?

This seemed to be a straightforward case of a Binomial distribution to me with:

n=30
r=1
p=0.02
q=0.98

Plugging into the Binomial formula gives me a probability of 0.333.

However, according to NASA it is 0.888.
And another text book gives 0.4545.

Can anyone explain the right answer?

Isn't your binomial result the probability of exactly 1 failure?

you could do a cumulative function but it's easier to do ControlG's method: prob 100% success = 0.98 ^ 30.

[DiscoStu]

Statistically the probability of failure would increase with each trip made. the more trips the higher the probability.

Taken as cumulative over the 30 flights, it would. However, we're presuming each flight is a discrete event.

[sasguru]

Isn't your binomial result the probability of exactly 1 failure?
you could do a cumulative function but it's easier to do ControlG's method: prob 100% success = 0.98 ^ 30.

Ah indeed that is the issue! Need more coffee methinks.
The NASA result is probably connected to the need for more funding ...

[DiscoStu]

If it requires 30 space shuttle flights to build the space station, and the probability of success for each flight is 98%, what is the probability of at least one failure in the sequence of 30 flights?

This seemed to be a straightforward case of a Binomial distribution to me with:

n=30
r=1
p=0.02
q=0.98

Plugging into the Binomial formula gives me a probability of 0.333.

However, according to NASA it is 0.888.
And another text book gives 0.4545.

Can anyone explain the right answer?

I wouldn't trust NASA, they can't tell the difference between centimetres and inches...