**Multinomial Experiment**

A **multinomial experiment **is an experiment that has the following properties:

- The experiment consists of
*n*repeated trials - Each trial has a discrete number of possible outcomes
- The probability that a particular outcome will occur on any given trial is constant
- Each trial is independent; the outcome of one trial does not affect the outcome of another

Suppose we toss a dice three times and record the number that the dice lands on each time. This is an example of a **multinomial experiment** because:

- The experiment consists of 3
- Each toss has a discrete number of possible outcomes (it can land on one of six numbers each time)
- The probability of any outcome is constant – the probabilities do not change from one toss to the next
- Each toss is independent; the outcome of one toss does not affect the outcome of another

*Note: A binomial experiment is a special case of a multinomial experiment in which each trial has exactly two possible outcomes.*

**Multinomial Distribution**

A **multinomial distribution **tells us the probability of getting a certain set of outcomes in a multinomial experiment.

Suppose we have a multinomial experiment with:

*n*trials*k*possible_{1}, O_{2}, … , O_{k})- each possible outcome can occur with probability p
_{1}, p_{2}, … , p_{k}

Then the probability (**P**) that O_{1 }occurs n_{1 }times, O_{2 }occurs n_{2 }times, … , and O_{k }occurs n_{k }times is:

**P** = [n! / (n_{1}! * n_{1}! * … * n_{k}! )] * (p_{1}^{n1} * p_{2}^{n2} * … * p_{k}^{nk})

Let’s walk through some examples to gain a better understanding of the multinomial distribution.

**Multinomial Distribution Example Problems**

**Example 1**

*We have 10 marbles in a bag – 2 green marbles, 3 blue marbles, and 5 yellow marbles. We randomly select 3 marbles from the bag, with replacement. What is the probability of selecting exactly 1 green, 1 blue, and 1 yellow marble?*

**Step 1: Identify the total number of trials ( n), the number of times we’re interested in selecting each possible color (n_{1}, n_{2}, … , n_{k}), and the probability of choosing each color on a given trial (p_{1}, p_{2}, … , p_{k}).**

*n *= 3 trials

*n _{i} *= we’re interested in the probability of selecting a green marble once, a blue marble once, and a yellow marble once, so n

_{1}= 1, n

_{2}= 1, n

_{3}= 1.

*p _{i} *= On any given trial, the probability of selecting a green marble is 2/10 = 0.2, the probability of selecting a blue marble is 3/10 = 0.3, and the probability of selecting a yellow marble is 5/10 = 0.5. Thus, p

_{1}= 0.2, p

_{2}= 0.3, p

_{3}= 0.5.

**Step 2: Plug these numbers into the multinomial formula or a multinomial calculator**

**Using the formula:**

**P** = [n! / (n_{1}! * n_{1}! * … * n_{k}! )] * (p_{1}^{n1} * p_{2}^{n2} * … * p_{k}^{nk})

**P** = [3! / (1! * 1! * 1! )] * (0.2^{1} * 0.3^{1 }* 0.5^{1})

**P** = **0.18**

If we randomly select three marbles from the bag with replacement, the probability that we select exactly one green marble, one blue marble, and one yellow marble is **0.18**.

**Using the calculator:**

Plug the following numbers into the Multinomial Distribution Calculator:

This matches the result that we got using the formula.

**Example 2**

*In a certain city, 20% of voters prefer candidate A, 40% prefer candidate B, and 40% prefer candidate C. If you randomly sample 10 voters, what is the probability that 2 will prefer candidate A, 4 will prefer candidate B, and 4 will prefer candidate C?*

**Step 1: Identify the total number of trials ( n), the number of voters we’re interested in preferring each possible candidate (n_{1}, n_{2}, … , n_{k}), and the probability of selecting a voter who prefers a certain candidate on a given trial (p_{1}, p_{2}, … , p_{k}).**

*n *= 10 trials

*n _{i} *= we’re interested in the probability of selecting 2 voters who prefer candidate A, 4 voters who prefe candidate B, and 4 voters who prefer candidate C, so n

_{1}= 2, n

_{2}= 4, n

_{3}= 4.

*p _{i} *= On any given trial, the probability of selecting a voter who prefers candidate A is 2/10 = 0.2, the probability of selecting a voter who prefers candidate B is 4/10 = 0.4, and the probability of selecting a voter who prefers candidate C is 4/10 = 0.4. Thus, p

_{1}= 0.2, p

_{2}= 0.4, p

_{3}= 0.4.

**Step 2: Plug these numbers into the multinomial formula or a multinomial calculator**

**Using the formula:**

**P** = [n! / (n_{1}! * n_{1}! * … * n_{k}! )] * (p_{1}^{n1} * p_{2}^{n2} * … * p_{k}^{nk})

**P** = [10! / (2! * 4! * 4! )] * (0.2^{2} * 0.4^{4 }* 0.4^{4})

**P** = **0.08258**

If we randomly select three voters, the probability that two of the voters prefer candidate A, four of the voters prefer candidate B, and four of the voters prefer candidate C is **0.08258**.

**Using the calculator:**

Plug the following numbers into the Multinomial Distribution Calculator:

This matches the result that we got using the formula.

**Example 3**

*We randomly select a card from a deck of cards 4 times without replacement. What is the probability that we select one spade, one club, one heart, and one diamond?*

**Step 1: Identify the total number of trials ( n), the number of times we’re interested in selecting each possible suit (n_{1}, n_{2}, … , n_{k}), and the probability of choosing each suit on a given trial (p_{1}, p_{2}, … , p_{k}).**

*n *= 4 trials

*n _{i} *= we’re interested in the probability of selecting each suit exactly one time, so n

_{1}= 1, n

_{2}= 1, n

_{3}= 1, and , n

_{4}= 1.

*p _{i} *= On any given trial, the probability of selecting a certain suit is 13/52 = 0.25. So, p

_{1}= 0.25, p

_{2}= 0.25, p

_{3}= 0.25, and p

_{4}= 0.25.

**Step 2: Plug these numbers into the multinomial formula or a multinomial calculator**

**Using the formula:**

**P** = [n! / (n_{1}! * n_{1}! * … * n_{k}! )] * (p_{1}^{n1} * p_{2}^{n2} * … * p_{k}^{nk})

**P** = [4! / (1! * 1! * 1! )] * (0.25^{1} * 0.25^{1 }* 0.25^{1 }* 0.25^{1})

**P** = **0.09375**

If we randomly select four cards from the deck with replacement, the probability that we select exactly one spade, one club, one heart, and one diamond is **0.09375**.

**Using the calculator:**

Plug the following numbers into the Multinomial Distribution Calculator:

This matches the result that we got using the formula.

**Example 4**

*A certain game consists of spinning a spinner, which can land on three different outcomes: a square, a circle, and a triangle. For any given spin, the probability of landing on a square is 0.7, a circle is 0.2, and a triangle is 0.1. If you randomly spin the spinner 10 times, what is the probability that it lands on a square 7 times, a circle 3 times, and a triangle 0 time**s?*

**Step 1: Identify the total number of trials ( n), the number of times we’re interested in landing on each possible outcome (n_{1}, n_{2}, … , n_{k}), and the probability of landing on each outcome on a given trial (p_{1}, p_{2}, … , p_{k}).**

*n *= 10 trials

*n _{i} *= we’re interested in the probability of landing on the square 7 times, the circle 3 times, and the triangle 0 times, so n

_{1}= 7, n

_{2}= 3, and n

_{3}= 0.

*p _{i} *= On any given trial, the probability of landing on a square is 0.7, the probability of landing on a circle is 0.2, and the probability of landing on a triangle is 0.1. So, p

_{1}= 0.7, p

_{2}= 0.2, and p

_{3}= 0.1.

**Step 2: Plug these numbers into the multinomial formula or a multinomial calculator**

**Using the formula:**

**P** = [n! / (n_{1}! * n_{1}! * … * n_{k}! )] * (p_{1}^{n1} * p_{2}^{n2} * … * p_{k}^{nk})

**P** = [10! / (7! * 3! * 0! )] * (0.7^{7} * 0.2^{3 }* 0.1^{0 })

**P** = **0.07906**

If we spin the spinner ten times, the probability that it lands on a square 7 times, a circle 3 times, and a triangle 0 times is **0.07906**.

**Using the calculator:**

Plug the following numbers into the Multinomial Distribution Calculator:

This matches the result that we got using the formula.