# Binomial, Poisson, and Multinomial Distributions

 Site: Saylor Academy Course: MA121: Introduction to Statistics Book: Binomial, Poisson, and Multinomial Distributions
 Printed by: Guest user Date: Tuesday, September 17, 2024, 2:03 PM

## Description

First, we will talk about binomial probabilities, how to compute their cumulatives, and the mean and standard deviation. Then, we will introduce the Poisson probability formula, define multinomial outcomes, and discuss how to compute probabilities by using the multinomial distribution.

##### Learning Objectives

1. Define binomial outcomes
2. Compute the probability of getting X successes in N trials
3. Compute cumulative binomial probabilities
4. Find the mean and standard deviation of a binomial distribution

When you flip a coin, there are two possible outcomes: heads and tails. Each outcome has a fixed probability, the same from trial to trial. In the case of coins, heads and tails each have the same probability of 1/2. More generally, there are situations in which the coin is biased, so that heads and tails have different probabilities. In the present section, we consider probability distributions for which there are just two possible outcomes with fixed probabilities summing to one. These distributions are called binomial distributions.

##### A Simple Example

The four possible outcomes that could occur if you flipped a coin twice are listed below in Table 1. Note that the four outcomes are equally likely: each has probability 1/4. To see this, note that the tosses of the coin are independent (neither affects the other). Hence, the probability of a head on Flip 1 and a head on Flip 2 is the product of P(H) and P(H), which is 1/2 x 1/2 = 1/4. The same calculation applies to the probability of a head on Flip 1 and a tail on Flip 2. Each is 1/2 x 1/2 = 1/4.

Table 1. Four Possible Outcomes.

The four possible outcomes can be classified in terms of the number of heads that come up. The number could be two (Outcome 1), one (Outcomes 2 and 3) or 0 (Outcome 4). The probabilities of these possibilities are shown in Table 2 and in Figure 1. Since two of the outcomes represent the case in which just one head appears in the two tosses, the probability of this event is equal to 1/4 + 1/4 = 1/2. Table 2 summarizes the situation.

 Number of Heads Probability 0 4-Jan 1 2-Jan 2 4-Jan

Figure 1. Probabilities of 0, 1, and 2 heads.

Figure 1 is a discrete probability distribution: It shows the probability for each of the values on the X-axis. Defining a head as a "success," Figure 1 shows the probability of 0, 1, and 2 successes for two trials (flips) for an event that has a probability of 0.5 of being a success on each trial. This makes Figure 1 an example of a binomial distribution.

Source: David M. Lane, https://onlinestatbook.com/2/probability/binomial.html
This work is in the Public Domain.

The binomial distribution consists of the probabilities of each of the possible numbers of successes on $N$ trials for independent events that each have a probability of п (the Greek letter pi) of occurring. For the coin flip example, $N=2$ and $\Pi=0.5$. The formula for the binomial distribution is shown below:

\begin{align*}P(x)=\frac{N !}{x !(N-x) !} \pi^{x}(1-\pi)^{N-x}\end{align*}

where $P(x)$ is the probability of $x$ successes out of $N$ trials, $N$ is the number of trials, and $п$ is the probability of success on a given trial. Applying this to the coin flip example,

\begin{align*}\begin{aligned}&P(0)=\frac{2 !}{0 !(2-0) !}\left(.5^{0}\right)(1-.5)^{2-0}=\frac{2}{2}(1)(.25)=0.25 \\&P(1)=\frac{2 !}{1 !(2-1) !}\left(.5^{1}\right)(1-.5)^{2-1}=\frac{2}{1}(.5)(.5)=0.50 \\&P(2)=\frac{2 !}{2 !(2-2) !}\left(.5^{2}\right)(1-.5)^{2-2}=\frac{2}{2}(.25)(1)=0.25\end{aligned}\end{align*}

If you flip a coin twice, what is the probability of getting one or more heads? Since the probability of getting exactly one head is $0.50$ and the probability of getting exactly two heads is $0.25$, the probability of getting one or more heads is $0.50+0.25=0.75$.

Now suppose that the coin is biased. The probability of heads is only $0.4$. What is the probability of getting heads at least once in two tosses? Substituting into the general formula above, you should obtain the answer .64.

We toss a coin 12 times. What is the probability that we get from 0 to 3 heads? The answer is found by computing the probability of exactly 0 heads, exactly 1 head, exactly 2 heads, and exactly 3 heads. The probability of getting from 0 to 3 heads is then the sum of these probabilities. The probabilities are: 0.0002, 0.0029, 0.0161, and 0.0537. The sum of the probabilities is 0.073. The calculation of cumulative binomial probabilities can be quite tedious. Therefore we have provided a binomial calculator to make it easy to calculate these probabilities.

Consider a coin-tossing experiment in which you tossed a coin 12 times and recorded the number of heads. If you performed this experiment over and over again, what would the mean number of heads be? On average, you would expect half the coin tosses to come up heads. Therefore the mean number of heads would be $6 .$ In general, the mean of a binomial distribution with parameters $\mathrm{N}$ (the number of trials) and $\sqcap$ (the probability of success on each trial) is:

\begin{align*}\mu=\mathrm{Nr}\end{align*}

where $\mu$ is the mean of the binomial distribution. The variance of the binomial distribution is:

\begin{align*}\sigma^{2}=N \pi(1-\pi)\end{align*}

where $\sigma^{2}$ is the variance of the binomial distribution.

Let's return to the coin-tossing experiment. The coin was tossed 12 times, so $N=12 .$ A coin has a probability of $0.5$ of coming up heads. Therefore, $\Pi=0.5$ The mean and variance can therefore be computed as follows:

\begin{align*}\begin{aligned}&\mu=\mathrm{Nm}=(12)(0.5)=6 \\&\sigma^{2}=\mathrm{Nm}(1-\pi)=(12)(0.5)(1.0-0.5)=3.0\end{aligned}\end{align*}

Naturally, the standard deviation $(\sigma)$ is the square root of the variance $\left(\sigma^{2}\right)$.

\begin{align*}\sigma=\sqrt{N \pi(1-\pi)}\end{align*}

Question 1 out of 6.
Select all that apply. Which of the following probabilities can be found using the binomial distribution?

The probability that 3 out of 8 tosses of a coin will result in heads

The probability that Susan will beat Shannon in two of their three tennis matches

The probability of rolling at least two 3's and two 4's out of twelve rolls of a die

The probability of getting a full house poker hand

The probability that all 5 of your randomly-chosen group members will have passed the midterm

The probability that a student blindly guessing will get at least 8 out of 10 multiple-choice questions correct

Question 2 out of 6.
You flip a fair coin 10 times. What is the probability of getting 8 or more heads?

Question 3 out of 6.
The probability that you will win a certain game is 0.3. If you play the game 20 times, what is the probability that you will win at least 8 times?

Question 4 out of 6.
The probability that you will win a certain game is 0.3. If you play the game 20 times, what is the probability that you will win 3 or fewer times?

Question 5 out of 6.
The probability that you will win a certain game is 0.3. You play the game 20 times. What is the mean of this binomial distribution?

Question 6 out of 6.
A biased coin has a .6 chance of coming up heads. You flip it 50 times. What is the variance of this distribution?

1. A binomial distribution has only two possible outcomes. You can think of them as successes and failures. For the correct answers, the successes are: a flip of heads, a win for Susan, a group member who has passed the midterm, and a correct answer on a multiple-choice question.

2. You may use the Binomial Calculator $(n = 10, p = .5, > or = 8)$. Otherwise add up the probability of getting 8, 9, and 10 heads: .044 + .01 + .001 = .055

3. Use the Binomial Calculator $(n = 20, p = .3, > or = 8). p = .23$

4. Use the Binomial Calculator ($n = 20, p = .3$, less than or $= 3)$. $p = .11$

5. $M = np = 20 \times .3 = 6$

6. $Var = np(1-p) = 50(.6)(1-.6) = 12$

The Poisson distribution can be used to calculate the probabilities of various numbers of "successes" based on the mean number of successes. In order to apply the Poisson distribution, the various events must be independent. Keep in mind that the term "success" does not really mean success in the traditional positive sense. It just means that the outcome in question occurs.

Suppose you knew that the mean number of calls to a fire station on a weekday is $8 .$ What is the probability that on a given weekday there would be 11 calls? This problem can be solved using the following formula based on the Poisson distribution:

$p=\frac{e^{-\mu} \mu^{x}}{x !}$ where

$e$ is the base of natural logarithms (2.7183)

$\mu$ is the mean number of "successes"

$\mathrm{x}$ is the number of "successes" in question

For this example,

\begin{align*}p = \frac{e^{-8} 8^{11}}{11 !} = 0.072\end{align*}

since the mean is 8 and the question pertains to 11 fires.

The mean of the Poisson distribution is $\mu$. The variance is also equal to $\mu$. Thus, for this example, both the mean and the variance are equal to 8.

Question 1 out of 1.
The mean number of defective products produced in a factory in one day is 21. What is the probability that in a given day there are exactly 12 defective products?

1. 0.012 can be obtained using the formula.

##### Learning Objectives
1. Define multinomial outcomes
2. Compute probabilities using the multinomial distribution
The binomial distribution allows one to compute the probability of obtaining a given number of binary outcomes. For example, it can be used to compute the probability of getting 6 heads out of 10 coin flips. The flip of a coin is a binary outcome because it has only two possible outcomes: heads and tails. The multinomial distribution can be used to compute the probabilities in situations in which there are more than two possible outcomes. For example, suppose that two chess players had played numerous games and it was determined that the probability that Player A would win is $0.40$, the probability that Player B would win is $0.35$, and the probability that the game would end in a draw is $0.25 .$ The multinomial distribution can be used to answer questions such as: "If these two chess players played 12 games, what is the probability that Player A would win 7 games, Player B would win 2 games, and the remaining 3 games would be drawn?" The following formula gives the probability of obtaining a specific set of outcomes when there are three possible outcomes for each event:

\begin{align*} \mathrm{p}=\frac{\mathrm{n} !}{\left(\mathrm{n}_{1} !\right)\left(\mathrm{n}_{2} !\right)\left(\mathrm{n}_{3} !\right)} \mathrm{p}_{1}^{\mathrm{n}_{1}} \mathrm{p}_{2}^{\mathrm{n}_{2}} \mathrm{p}_{3}^{\mathrm{n}_{3}} \end{align*}

where

$\mathrm{p}$ is the probability,

$\mathrm{n}$ is the total number of events

$\mathrm{n}_{1}$ is the number of times outcome 1 occurs,

$\mathrm{n}_{2}$ is the number of times outcome 2 occurs,

$\mathrm{n}_{3}$ is the number of times outcome 3 occurs,

$\mathrm{p}_{1}$ is the probability of outcome 1

$\mathrm{p}_{2}$ is the probability of outcome 2 , and

$\mathrm{p}_{3}$ is the probability of outcome 3.

For the chess example,

$\mathrm{n}=12$ (12 games are played),

$\mathrm{n}_{1}=7$ (number won by Player A),

$\mathrm{n}_{2}=2$ (number won by Player B),

$\mathrm{n}_{3}=3$ (the number drawn),

$\mathrm{p}_{1}=0.40$ (probability Player A wins)

$\mathrm{p}_{2}=0.35$ (probability Player B wins)

$\mathrm{p}_{3}=0.25$ (probability of a draw)

$\mathrm{p}=\frac{12 !}{(7 !)(2 !)(3 !)} \cdot 40^{7} \cdot 35^{2} \cdot 25^{3}=0.0248$

The formula for $\mathrm{k}$ outcomes is

\begin{align*} \mathrm{p}=\frac{\mathrm{n} !}{\left(\mathrm{n}_{1} !\right)\left(\mathrm{n}_{2} !\right) \ldots\left(\mathrm{n}_{\mathrm{k}} !\right)} \mathrm{p}_{1}^{\mathrm{n} 1} \mathrm{p}_{2}^{\mathrm{n}_{2}} \ldots \mathrm{p}_{\mathrm{k}}^{\mathrm{n}} \end{align*}

Note that the binomial distribution is a special case of the multinomial when $\mathrm{k}=$ $2$.

Question 1 out of 1.
In a certain town, 40% of the eligible voters prefer candidate A, 10% prefer candidate B, and the remaining 50% have no preference. You randomly sample 10 eligible voters. What is the probability that 4 will prefer candidate A, 1 will prefer candidate B, and the remaining 5 will have no preference?