What is the Probability Function for Tossing a Coin 10 Times?

[1MileCrash]

Suppose that a fiat coin is tossed 10 times independently. Determine the pf of the number of heads that will be obtained.

I'm feeling really out of touch with this material. It seems like any number of heads is equally likely but I know that just can't be right.

My reasoning for saying such is that the probability of a head is 1/2 and the probability of a tail is 1/2,

So the probability of a head, then tail, then head is

(1/2)(1/2)(1/2)

Or head, head, head is

(1/2)(1/2)(1/2)

But I know that getting 5 heads in the ten flips and getting all heads aren't equally likely.. are they?

So can someone "remind" me how my thinking is wrong?

 

[1MileCrash]

Wait, I think I see. The probability of any order of heads or tails is equally likely, but the number of "ways" 5 heads can occur is more than the number of ways 10 can occur (which is just one.)

So

f(x) = (10 CHOOSE x)(1/2)^10

Right?

 

[LCKurtz]

Yes. You might want to review the binomial distribution.

 

[1MileCrash]

Indeed. Should I also make the function piecewise, 0 for any exception to the integer interval 1 through 10?

 

[bossman27]

Since each trail can be classified as a simple success or failure, we use the binomial distribution, as you have already written down for this case. Just incase, the general form is:

$P(X) = C^{n}_{x}p^{x}q^{n-x}$

where:
p = probability of success in a trial
q = probability of failure = (1-p)

Edit: Yes, the distribution is inherently discontinuous at non-integer values since there can only be an integer number of successes in n trails.