# The probability for a range of numbers

• Oct 16th 2012, 12:49 PM
TheMegaSage
The probability for a range of numbers
Let's say you have 24 people who are tossing a penny into a can. Each toss has a straight 15% chance of success. How do I figure out what percentage chance for each number, from 0 to 24, has of being the total number of successful throws on average?

Basically I'd like to see that there's a 2% chance that 0 throws are successful (made up number!), a 5.5% chance (made up number!!) that 4 throws are successful, down to a 0.0001% chance (made up number!!!) that all 24 throws are successful. And is there a way to mathematically 'phrase' that in such a way as to say "the likelihood is that between 3 and 5 throws are successful, on average"?

Any help would be appreciated. I'm not particularly proficient with math, so any other ways to consider this data would also be interesting! :)

• Oct 16th 2012, 12:55 PM
Plato
Re: The probability for a range of numbers
Quote:

Originally Posted by TheMegaSage
Let's say you have 24 people who are tossing a penny into a can. Each toss has a straight 15% chance of success. How do I figure out what percentage chance for each number, from 0 to 24, has of being the total number of successful throws on average?

$\mathcal{P}(k)=\binom{24}{k}(0.15)^k(0.85)^{24-k},~k=0,1,\cdots,24$
• Oct 16th 2012, 12:59 PM
ebaines
Re: The probability for a range of numbers
This is called binomial probability and there's a straight-forward formula for this. If the probability of a successful toss is 'p,' then the probability of having 'k' successful tosses out of 'n' attempts is:

$p(k) = C(N,K) p^k(1-p)^{n-k}$

where $C(n,k)$ means the combinations of N things taken k at a time, and is equal to:

$C(n,k) = \frac {n!} {(n-k)!k!}$

To use your example: if p=15%, the chance of getting 0 out of 24 tosses in the can is $C(24,0)0.15^0 0.85^{24} = 2\%$ (approximately). The chance of getting 5 is $C(24,5)0.25^5 0.85^{19} = 14.7\%$. And the chance of getting all 24 in is $0.15^{24} = 1.68 x 10^{-20}$, or about 1 in 600 quintillion!
• Oct 16th 2012, 01:09 PM
TheMegaSage
Re: The probability for a range of numbers
That is fantastic! Thank you both so much.

Here's the formula in Excel:
=COMBIN(24,A1)*(0.15)^A1*(0.85)^(24-A1)

Where A1 = 0, B1 = A1+1, and so on.

And just to close off this fun little exercise, here are the results from 0 to 24. :)

0 2.02327174731858000000%
1 8.56915092981985000000%
2 17.39033571051670000000%
3 22.50514033125700000000%
4 20.85035060101750000000%
5 14.71789454189470000000%
6 8.22470577341175000000%
7 3.73221942659020000000%
8 1.39958228497133000000%
9 0.43908463842237700000%
10 0.11622828664121700000%
11 0.02610474887128950000%
12 0.00499061375480534000%
13 0.00081295065689136800%
14 0.00011272004906476900%
15 0.00001326118224291410%
16 0.00000131636735499515%
17 0.00000010931770422105%
18 0.00000000750219538772%
19 0.00000000041807899993%
20 0.00000000001844466176%
21 0.00000000000061998863%
22 0.00000000000001491951%
23 0.00000000000000022894%
24 0.00000000000000000168%