Probability problem for statisticians/mathematicians

[BrilloPad]

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?

Not sure about Nasa(got a link?). But I think the text book is right.

http://en.wikipedia.org/wiki/Binomial_distribution

n=30
k=0
p=0.02 i.e. success=failure!

Pr(K=0) = binomial(n,k) * power(p,k)*power(1-p, n-k)

= 1 * 1 * power(0.98, 30) = 0.545484319

So failure is not getting the above = 0.4545

Where did you get your formula from?

[dmini]

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

No it doesnt. Thats where a lot go wrong. Probability is reset on each and every turn.
eg
If you toss a coin, it doesnt matter if the last 50 have all been heads - the probability of the next one being a head (or a tail) is still .5

(dmini in her Pure Maths & stats degree mode!)

[BrilloPad]

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

But they got a man to the moon. Or was it to a TV studio in Vegas?

[bobhope]

yeah 0.333 is the probability that exactly one flight will go wrong

30 * (0.02 * 0.98^29) = 0.333969991

[Marina]

No it doesnt. Thats where a lot go wrong. Probability is reset on each and every turn.
eg
If you toss a coin, it doesnt matter if the last 50 have all been heads - the probability of the next one being a head (or a tail) is still .5

(dmini in her Pure Maths & stats degree mode!)

A space shuttle is a lot more complicated than a coin. If you applied that logic to a car, you'd never need to have it serviced (although I assume shuttles are intensively serviced between each trip, but even so they must get more unreliable eventually just the same as planes)

[BrilloPad]

A space shuttle is a lot more complicated than a coin. If you applied that logic to a car, you'd never need to have it serviced (although I assume shuttles are intensively serviced between each trip, but even so they must get more unreliable eventually just the same as planes)

Maybe that is the difference? Nasa assumes it gets more unreliable. But that was not the question posed.

[sasguru]

Not sure about Nasa(got a link?). But I think the text book is right.

http://en.wikipedia.org/wiki/Binomial_distribution

n=30
k=0
p=0.02 i.e. success=failure!

Pr(K=0) = binomial(n,k) * power(p,k)*power(1-p, n-k)

= 1 * 1 * power(0.98, 30) = 0.545484319

So failure is not getting the above = 0.4545

Where did you get your formula from?

Used same formula, but got the answer for exactly one failure not at least one failure. See Bobhope's answer above. Your calculation is correct.
Thanks to all who replied - I'll have more problems for you soon

[Bagpuss]

You must be bored if you're doing a/s level maths.
I hope this isn't your coursework!

[Lucy]

You must be bored if you're doing a/s level maths.
I hope this isn't your coursework!

[HairyArsedBloke]

In reliability problems such as this, aren’t you supposed to use the Poisson distribution?