Using NProbability[] or Probability[] to work out the probability of 3 or more Heads from 4 coin tosses

Question

Is it possible to work out the probability of 3 or more Head from 4 coin tosses using the Probability or NProbability functions.

This is not a question about the trivial answer to this problem, it is more to get an understanding of how to solve this kind of problem with Mathematica using distributions.

So using 4 random variables from Distribution P

I was hoping something like this would do the trick, but it does not work. I get 0.

P = BernoulliDistribution[0.5];
vars = List[Distributed[a,P],Distributed[b,P],Distributed[c,P],Distributed[c,P]];
NProbability[Count[ {a,b,c,d}, 1] >= 3,  vars]

Any ideas would be greatly appreciated.

Banta

Two errors there: c declared twice and d never and using Count instead of Total. – Featherhead

Perhaps @Sjoerd is being too polite. After his modifications the OP's code works – Curvet

Ha! So it does. The double declaration was a copy paste error to the site. Using Total instead of Count did the trick. Many thanks guys. – Banta

Answer 1

Not an expert using Mma for statistics here, but this seems to work:

l = TransformedDistribution[
       x + y + w + z, {x \[Distributed] BernoulliDistribution[0.5], 
                       y \[Distributed] BernoulliDistribution[0.5], 
                       z \[Distributed] BernoulliDistribution[0.5], 
                       w \[Distributed] BernoulliDistribution[0.5]}];

Table[NProbability[x > i, x \[Distributed] l], {i, -1, 4}]
(*
{1, 0.9375, 0.6875, 0.3125, 0.0625, 0.}
*)

Curvet

Wow. Thanks so much! Explained it perfectly :) – Banta

At the mma tech conference right now with leonid. We don't have much tine doing anything else. – Featherhead

@Sjoerd Conference? With or without beer? – Curvet

We had a social yesterday where we could try the products of local brewery. Leonid made us miss the last bus back, but Daniel was kind enough to being us back safe. – Featherhead

Of course, we talked a lot about you and mr wizard. You were missed. – Featherhead

Lots of stuff under NDA. Can't tell much. – Featherhead

@Sjoerd Enjoy! And upload some photographs! – Curvet

Answer 2

In[10]:= Probability[a + b + c + d >= 3, vars]

Out[10]= 0.3125

Coin flipping is easier described with a BinomialDistribution:

In[12]:= Probability[m >= 3, m \[Distributed] BinomialDistribution[4, 0.5]]

Out[12]= 0.3125

Featherhead

Yes that is a lot cleaner. I did notice the TransformedDistribution as suggested by belisarius gets evaluated to BinomialDistribution. Thankyou :) – Banta

@Banta I thought you did that on purpose because you wanted to know how to add random vars! Ha! Silly me. – Curvet

@belisarius I did want to know how to add random vars and not just this yoy coin flip problem. So it being evaluated to a BinomialDist is just an interesting point in this particular case. What you provided is what really I wanted, a general understanding of how do these kinds of calculations. So you were spot on! :) – Banta