Probability of the number of heads given a number of coin flips

Question

I've been learning statistics through the website Khan Academy and they have this problem:

Let the random variable X = the # of heads from flipping a coin 5 times. The total number of outcomes from 5 flips = 32. Figure out the probability that the random variable takes on 0, 1, 2, 3, 4, or 5. The instructor then answers the problem by giving the probabilities below:

P(X = 0) = 1/32

P(X = 1) = 5/32

P(X = 2) = 10/32

P(X = 3) = 10/32

P(X = 4) = 5/32

P(X = 5) = 1/32

The way he got the probabilities/answers is through using combinations: $$\frac{_{5}C_X}{32} $$

I've spent so much time trying to find out why the instructor uses combinations here. Can someone please explain to me?

Answer 1

The definition of the combinations function $_nC_k$ is the number of ways to choose $k$ objects from a set of $n$ objects. To find the number of ways to flip $x$ heads when you flip $5$ coins, you must sum the probabilities of each different way of doing this. There are $_5C_x$ ways to “choose” which $x$ coins come up heads. For example, for $x=2$: $$HHTTT$$ $$HTHTT$$ $$HTTHT$$ $$HTTTH$$ $$THHTT$$ $$THTHT$$ $$THTTH$$ $$TTHHT$$ $$TTHTH$$ $$TTTHH$$ There are $_5 C_2=10$ ways to do this, and the probability of each option is $1/2^5$, so the total probability of getting $2$ heads is $_5 C_2/2^5$. More generally, the probability of flipping $x$ heads is $$\frac{_5 C_x}{2^5}$$

Answer 2

A more complicated way to look at this is to realize that X is a binomial random variable, with probability of success (head) equal to $p =0.5$ and number of trials (toin cosses) $n=5$. If, for example, you compute $P(X=1)$ for this binomial distribution using one of the calculators available online, you will obtain 0.3125, that is, $5/32$.