Summary

This is a “Rpub format” of my page: https://dataz4s.com/statistics/the-binomial-distribution/

Four properties of the binomial distribution

1. Fixed number of trials (n)
2. Two outcomes in a trial, success or failure
3. Trials are independent
4. Probability of success (p) remains constant

Explained with these slides:

Calculations in the binomial distribution

Let X be the number of tails from flipping a fair coin 5 times. We can denote:

\[X= \text{the number of tails in 5 flips}\] \[\Leftrightarrow X=0,1,2,3,4,5\] \[X\sim B(5; 0.5)\]

Total possible outcomes = 32: \[2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32\] The probability of flipping 0 tails in 5 flips can be written: \[\frac{^5C_0}{32}\]

Each point probability for flipping 0-5 tails in 5 flips can be expressed like this:

\[P(X=0)= \frac{^5C_0}{32}= \frac{1}{32}\] \[P(X=1)= \frac{^5C_1}{32}= \frac{5}{32}\] \[P(X=2)= \frac{^5C_2}{32}= \frac{10}{32}\] \[P(X=3)= \frac{^5C_3}{32}= \frac{10}{32}\] \[P(X=4)= \frac{^5C_4}{32}= \frac{5}{32}\] \[P(X=5)= \frac{^5C_5}{32}= \frac{1}{32}\]

If I have a 25% chance of winning a tennis match, what is then the probabiltiy that I win exactly 4 matches out of 6?

\[P(\text{Exactly 4 matches out of 6}) = \] \[P(X=4) = \binom{6}{4}\times 0.25^4 \times 0.75^2 \]

There are 15 possible outcomes of the 6 matches:

\[^6C_4 = \binom{6}{4} = \frac{6!}{4!(6-4)!} = 15\]

The point probabilities:

\[P(\text {k wins in n attempts}) = \binom{n}{k}\times \text {win}^{k} \times lose^{n-k} \]

Generalized formula:

\[\Leftrightarrow P(k) = \binom{n}{k}\times p^{k} \times (1-p)^{n-k} \]

What is my probability of losing more than 3 times out of 6 matches?

\[\text{X= Number of lost matches}\] \[X\sim B(n=6;p=0.75)\] \[P(X>3)\]

\[P(X=0) = \binom{6}{0}\times p^{0} \times (1-0.75)^{6-0} = 0.0002\] \[P(X=1) = \binom{6}{1}\times p^{1} \times (1-0.75)^{6-1}= 0.0044\] \[P(X=2) = \binom{6}{2}\times p^{2} \times (1-0.75)^{6-2}=0.0330\] \[P(X=3) = \binom{6}{3}\times p^{3} \times (1-0.75)^{6-3}=0.1318 \] \[P(X=4) = \binom{6}{4}\times p^{4} \times (1-0.75)^{6-4}=0.2966 \] \[P(X=5) = \binom{6}{5}\times p^{5} \times (1-0.75)^{6-5}= 0.3560\] \[P(X=6) = \binom{6}{6}\times p^{6} \times (1-0.75)^{6-6}= 0.1780\] \[\Leftrightarrow P(X>3) = 0.2966 + 0.3560 + 0.1780 = 0.8306\]

Binomial distribution with R

Let’s look at the dbinom, pbinom, rbinom and qbinom functions in R:

Say that X is binomially distributed with n=20 trials and p=1/6 prob of success:

\[X\sim BIN(n=20; p=1/6)\]

# P(X=3)
dbinom(x=3, size = 20, prob = 1/6)

## [1] 0.2378866

# P(X=0) & P(X=1) & P(X=2) & P(X=3)
dbinom(x=0:3, size = 20, prob=1/6)

## [1] 0.02608405 0.10433621 0.19823881 0.23788657

# P( <= 3)
sum( dbinom(x=0:3, size = 20, prob=1/6) )

## [1] 0.5665456

# P( <= 3)
pbinom(q=3, size = 20, prob = 1/6, lower.tail = T)

## [1] 0.5665456

rbinom(5,150,0.05)

## [1] 12  5  6  8  8

# 10 coin flips
rbinom(10, 1,.5)

##  [1] 1 0 1 0 0 0 0 0 0 0

# 10 flips of 100 coins
rbinom(10,100,.5)

##  [1] 52 47 52 55 49 47 48 47 46 50

qbinom(0.25,10,.5)

## [1] 4