Binomial Distribution

The binomial distribution applies to events that can be described as a "success" if one outcome occurs or a "failure" if any other outcome occurs. There can be more than 2 outcomes, but it needs to be black and white in terms of success or failure.

The binomial distribution is given by the equation

(1)
\begin{align} P(X=r)= \left ( \begin{matrix}n\\ r\end{matrix}\right ) (1-p)^{n-r}p^r \end{align}

Where:

  • p is the probability that an event is successful (i.e. the probability of getting heads)
  • 1-p represent the probability that the event is a failure
  • r is the number of successful outcomes
  • n is the number of trials or attempts.

For example if we want to know the probability of getting 3 heads on a coin flip when flipping the coin a total of 5 times then:

(2)
\begin{align} P(X=3)= \left ( \begin{matrix}5\\ 3\end{matrix}\right ) (1-0.5)^{5-3}(0.5)^3 \end{align}

The coefficient $\left ( \begin{matrix}5\\ 3\end{matrix}\right )$ can be calculated with your calculator, see the page on Binomial Theorem.

Connection to Binomial Theorem

Yes, there is a remarkable similarity between a binomial distribution and a binomial expansion… Click below for more details.

We'll build this up with an example. If I flip the coin once there are two possible outcomes: H and T. Each outcome occurs 1 time.

If I flip a coin twice there are 4 possible outcomes HH, TH, HT and TT. Notice that the outcome of 1 head and 1 tail occurs twice (or two different ways). Putting this into a table:

Flips Outcomes
1 H,T
2 HH, TH, HT,TT
3 HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

For three flips of a coin there are 4 separate outcomes 3 Heads (1 occurrence), 2 Heads and 1 Tail (3 occurrences), 1 Head and 2 Tail (3 occurrences) and 3 Tails (1 occurrence).

Now if we make a new table with the number of occurrences of each outcome we get something like:

Flips Occurrences
1 1 1
2 1 2 1
3 1 3 3 1

This pattern should look familiar. Its the 2nd through 4th row of Pascals triangle, which is the pattern that binomial expansions follow! Lets expand the table bigger to try to explain a bit more.

Flips Occurrences
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1

If we look at the last row this represents flipping a coin 5 times. This shows us that there are only 1 way of getting 5 heads, 5 ways of getting 4 heads and 1 tails, 10 ways of getting 3 heads and 2 tails, etc…

This is the very same idea as with binomial expansions!

Example 1

If we want to know the probability of a certain outcome, say 4 heads out of 5 flips, first we need to calculate the probability this occurring:

(3)
\begin{align} \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{32} \end{align}

BUT! This can happen 5 different ways for we must multiply by 5 giving a final probability of: $\frac{5}{32}$.

Using the formula from above:

(4)
\begin{align} P(X=4)= \left ( \begin{matrix}5\\ 4\end{matrix}\right ) (1-0.5)^{5-4}(0.5)^4 \end{align}
Example 2

Find the probability of rolling a number less than 5 on a die for 3 out of 4 rolls.

First the probability of rolling less than a 5 is $\frac{4}{6}$ and the probability of rolling 5 or more is $\frac {2}{6}$. So the total probability is:

(5)
\begin{align} \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{4}{6} \cdot \frac{2}{6} = \frac{128}{1296} \end{align}

But, there are 4 different ways for this to occur (SSSF, SSFS, SFSS, FSSS - S is for success and F is for failure) so we have to multiply our probability by 4.

(6)
\begin{align} P(X=3)=\frac{512}{1296} \end{align}

Using the equation from above:

(7)
\begin{align} P(X=3)= \left ( \begin{matrix}4\\ 3\end{matrix}\right ) (1-\frac{2}{6})^{4-3}(\frac{4}{6})^3 \end{align}

Okay, but the 1st Row of Pascal's Triangle is Missing!

This corresponds to $n=0$ and $r=0$. Which in the language of probability means we did zero trials ($n=0$) to which there is only 1 outcome (the value in the 1st row) and that is zero successes ($r=0$).


Want to add to or make a comment on these notes? Do it below.

inkognito (guest) 29 Mar 2012 21:06

Why is the probability of a successful event 0.5? If you want the probability of getting 3 heads when flipping the coin 5 times, isn't the probability of success 3 out of 5 (which is 0.6)?

by inkognito (guest), 29 Mar 2012 21:06
Hank E Stevens Hank E Stevens 30 Mar 2012 03:43

The "successful event" is getting heads once so P(heads)= 0.5 Another way to think of this is each event is separate or independent, i.e. each flip of the coin does not affect the next.

In order to find the probability of getting EXACTLY 3 heads out of 5 we need to consider all the different ways it can be done.

HHHTT
HHTHT
HHTTH

etc…

The coefficient on equation (1) gives us the number of ways it can happen dependent on n and r.

Note that the probability of HHHTT is exactly the same as HHTHT (or for that matter HHHHH). That is P(HHHTT) = (0.5)(0.5)(0.5)(0.5)(0.5)=0.031250 However, there is only one way to get HHHHH which makes it "rare" whereas there are many ways to get 3 heads out of 5.

Hope that helps.

by Hank E Stevens Hank E Stevens , 30 Mar 2012 03:43
mathsy (guest) 12 Apr 2015 16:04

the probability of a successful event is the probability of getting heads, which on a two sided coin, is one in two —> 0.5

the 3 times makes no relevance to the probability of getting a head, since doing it three times doesn't affect the outcome of the coin being heads or tails

by mathsy (guest), 12 Apr 2015 16:04
Sachin (guest) 03 Dec 2014 12:39

Line marked (7) above is incorrect. Probability of success and failures are swapped in the formula

by Sachin (guest), 03 Dec 2014 12:39

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page revision: 15, last edited: 18 Feb 2013 17:29
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