# Thread: The probability for a range of numbers

1. ## 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!

Thanks for your help.

2. ## Re: The probability for a range of numbers

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?
$\displaystyle \mathcal{P}(k)=\binom{24}{k}(0.15)^k(0.85)^{24-k},~k=0,1,\cdots,24$

3. ## 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:

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

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

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

To use your example: if p=15%, the chance of getting 0 out of 24 tosses in the can is $\displaystyle C(24,0)0.15^0 0.85^{24} = 2\%$ (approximately). The chance of getting 5 is $\displaystyle C(24,5)0.25^5 0.85^{19} = 14.7\%$. And the chance of getting all 24 in is $\displaystyle 0.15^{24} = 1.68 x 10^{-20}$, or about 1 in 600 quintillion!

4. ## 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%