Probability problem

p ≈ 0.0414

$${\left({\frac{{\mathtt{20}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{0.041\: \!4}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{0.041\: \!4}}\right)}^{{\mathtt{15}}} = {\mathtt{0.001\: \!000\: \!015\: \!738\: \!633\: \!1}}$$

So the overall probability should be about 4%.

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I am not at all sure that I understand the question but the probability of exactly five being broken is given by this (I think)  Where p is the prob of an individual one being broken.

In my interpretation of the question (which could easily be totally incorrect) the 400 and the 8% are irrelevant.

The web 2 calc won't solve this.  Maybe the wolfram|alpha calc will give you an answer.

$${\left({\frac{{\mathtt{20}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{20}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)}{\mathtt{\,\times\,}}{{\mathtt{p}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{\left({\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{p}}\right)}^{{\mathtt{15}}} = {\mathtt{0.001}} = \tiny\text{Error: }$$

If you get the proper answer I'd really like you to post it.

Any one else here want to give it a shot?

How did you get the 0.0414 Alan.  

I assume you used a different calc?

Melody, WolframAlpha will solve this.............http://www.wolframalpha.com/input/?i=solve+C%2820%2C5%29+*p^5*%281-p%29^15+%3D+.001

thanks Chris, i thought wolfram would give the answer.  I just had not gone there.  :)

What I did was to divide both sides by ncr(20,5) and then take the fifth root of both sides.  This leaves a quartic in p.  There are four solutions to this quartic; two real and two complex.  The 0.0414 was one of the real solutions (the other was around 0.6, so I doubt this was the desired result!).

(I used Mathcad to do the calculations).