# Thread: Finding probability functions

1. ## Finding probability functions

Four fair coins are tossed simultaneously. Find the probability of the random variable X = number of heads and compute the following probabilities:
a) obtaining no heads
b) precisely 1 head
c) at least 1 head
d) not more than 3 heads.

I'm not sure how to set up these probabilities.

For example problems, the book has $f(x) = \begin{pmatrix}n\\ x \end{pmatrix}p^xq^{n-x}$ for a binomial distribution and

$f(x) = \begin{pmatrix}n\\ x\end{pmatrix}(\frac{1}{2})^n$ for a symmetric case. There are other probability functions that they list as well.

My biggest problem is finding out the proper way to write these probabilities out, or what probability function can be used (if any).

For example, in a), I can logically see that it would be (1/2)*(1/2)*(1/2)*(1/2), but how can I write this as a probability function?

2. ## Re: Finding probability functions

Originally Posted by deezy
Four fair coins are tossed simultaneously. Find the probability of the random variable X = number of heads and compute the following probabilities:
a) obtaining no heads
b) precisely 1 head
c) at least 1 head
d) not more than 3 heads.

I'm not sure how to set up these probabilities.

For example problems, the book has $f(x) = \begin{pmatrix}n\\ x \end{pmatrix}p^xq^{n-x}$ for a binomial distribution and

$f(x) = \begin{pmatrix}n\\ x\end{pmatrix}(\frac{1}{2})^n$ for a symmetric case. There are other probability functions that they list as well.

My biggest problem is finding out the proper way to write these probabilities out, or what probability function can be used (if any).

For example, in a), I can logically see that it would be (1/2)*(1/2)*(1/2)*(1/2), but hibution B(ow can I write this as a probability function?
The number of heads has a binomial distribution $\text{B}(4,\ 0.5)$, so the probability of $r$ heads is:

$p(numb\_ heads=r)=b(n;4,\ 0.5)= \begin{pmatrix}4\\ r \end{pmatrix}(0.5)^r (0.5)^{4-r}=\frac{4!}{(4-r)! r!}(0.5)^4$

So if $r=0$ we have:

$p(numb\_ heads=0)=\frac{4!}{(4-0)! 0!}(0.5)^4=(0.5)^4$

CB

3. ## Re: Finding probability functions

Not sure how to write c).

I've tried doing the probability of 1 head, 2 heads, 3 heads, and 4 heads separately and adding/multiplying them together.

Is it correct to do these probabilities separately?

4. ## Re: Finding probability functions

Originally Posted by deezy
Not sure how to write c).
At least one is the complement of none.
$1-P(\text{none}).$