# What is the probability that a fair coin comes up tail three out of four flips?

As the question asks, what is the probability that a fair coin comes up tail three out of four flips? I know the probability of getting tails on one flip is 1/2, but I'm not sure how to solve this for three/four flips. I was thinking about using a permutation of P(4,3). I don't have to do out all the arithmetic.

A $5^{\text{th}}$ graders attempt at your question :
$$\mathrm { \color{red}{HHHH},\\ \color{red}{HHH}\color{blue}{T}, \color{red}{HH}\color{blue}{T}\color{red}{H}, \color{red}{H}\color{blue}{T}\color{red}{HH}, \color{blue}{T}\color{red}{HHH},\\ \color{red}{HH}\color{blue}{TT}, \color{blue}{T}\color{red}{HH}\color{blue}{T}, \color{blue}{TT}\color{red}{HH}, \color{red}{H}\color{blue}{TT}\color{red}{H}, \color{red}{H}\color{blue}{T}\color{red}{H}\color{blue}{T}, \color{blue}{T}\color{red}{H}\color{blue}{T}\color{red}{H},\\ \boxed{\color{red}{H}\color{blue}{TTT}, \color{blue}{T}\color{red}{H}\color{blue}{TT}, \color{blue}{TT}\color{red}{H}\color{blue}{T}, \color{blue}{TTT}\color{red}{H}}\ ,\\ \color{blue}{TTTT}}$$

Of the $2^4$ paths in the sample space, there are $4$ paths containing $\color{blue}{\mathrm T}$ thrice.

Hence, you have a $\dfrac{4}{2^4} = \dfrac{1}{4} = 25\%$ chance of landing a tail $3$ times out of $4$ coin flips.

The number of different outcomes is $2^4 = 16$. There are 4 outcomes that give three heads (easily seen that the first, second, third or fourth flip must be tails with all the rest heads), so the probability for three heads is 4/16 = 1/4.

More generally, in n flips there are $2^n$ outcomes, and for $m \le n$ there are $^nC_m = n!/m!(n-m)!$ ways to get m heads, giving a probability of $n!/m!(n-m)!2^n$

• I'm a little confused about where the number of outcomes that give three heads comes from. What if I were to say four out of five flips must be heads? Thanks by the way, this is helpful. – Carl Nov 20 '14 at 14:25
• Nevermind, I get it now. Thanks – Carl Nov 20 '14 at 14:31

You note that for independent events $A_{i}$ we have the formula:

$$P\left(\bigcap_{i}A_{i}\right)=\prod_{i}P(A_{i})\tag{1}$$

Which can loosely be read as the probability of all of the $A_{i}$ independent events happening is equivalent to the product of their individual probabilities of happening.

Applying this to your problem, we have 4 independent coin tosses, and so the probability of only one heads appearing is:

$$P(\text{only one H}) = P(\text{TTTH})+P(\text{TTHT})+P(\text{THTT})+P(\text{HTTT})$$

But we note that $P(TTTH)=P(TTHT)=P(THTT)=P(HTTT)$ as they all consist of probabilities of a set of independent events all occuring, so we have:

$$P(\text{only one H})=4P(\text{TTTH})=4P(T)P(T)P(T)P(H)=4\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)=\frac{1}{4}$$

We can also note that in general this leads us to a probability distribution called the Geometric distribution, which is the probability of $n-1$ failures followed by a success where the probability of success is a constant $p$.

We have:

$$P(n)=P(n-1 \text{ failures followed by a success})=P\left(\bigcap_{i=1}^{n-1}F_{i}\cap S\right)$$

Where $F_{i}$ are the $i$th failure event and $S$ is the success event. We can use $(1)$ to thus write:

$$P(n)=\prod_{i=1}^{n-1}(1-p) \times p=p(1-p)^{n-1}$$