Probability Binomial Distribution

Hi,
This isn't actually a programming question, but I figure most people reading this part of the forum are good at maths so here it is:

Russel Narks is shooting at a target. His probability of hitting the target is 0.6. What is the minimum number of shots needed for the probability of Russel hitting the target exactly five times to be more than 25%.
This is a binomial distribution equation, which for those of you that remember binomial distribution has the formula:
Pr(X=x)= (n C x) * (p^x) * ((1-p)^(n-x)
Where n = number of trials
p = chance of success
and x = number of successes.
So anyway with the above equation I got

.25 = (n C 5) * (.6^5) * (.4^(n-5))
I then rearranged it to
.25/(.6^5) = (n C 5) * (.4^(n-5))

and then got stuck there...if anyone has any ideas please tell me...(I could had gone about this the total wrong way so my workings may be useless)

Can you rememebr the formula for n C 5 it is something like
(n-5)!*5!/n! If you can dig that out you can render the equation down to just n and numbers...

hth

Cheers,

P.

Quote:

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Can you rememebr the formula for n C 5 it is something like
(n-5)!*5!/n!

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Nearly, it should be:
(n C x) = n!/(n-x)!*x!

You should find that 7 attempts are needed.

The formula you posted looks correct for the probability of exactly five hits in N tries.

Using this formula results in misleading conclusions. You get the following probabilites (Multiply by 100 for chances expressed in percentages).

.186624 for N = 6
.261274 for N = 7
.278692 for N = 8
.250823 for N = 9
.200658 for N = 10

From 10 on, the probability of exactly 5 hits decreases and approaches zero as N grows without bound.

I think you should be asking a slightly different question.

Perhaps you should ask how many shots are required to result in a 25% (or better) chance of at least five hits. I got the following for this question.

.233280 for N = 6
.419904 for N = 7
.594086 for N = 8

The above indicates that the chances of 5 hits gets better as N increases.

I do not guarantee the above numbers, but am fairly sure they are correct. I have an HP programmable calculator which I used for the above.

Damn! Nearly right - I must have had the Australian method :)

Still I did that 14 years ago, so it wasn't bad...

P.