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3.3: A Glimpse at Binomial and Multinomial Distributions

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In Chapter 4, we will learn more about binomial and multinomial distributions. However, we are now talking about probability distributions, and as such, we should at least see how the problems change for these distributions. We will briefly introduce the concepts and their formulas here, and then we will get into more detail in Chapter 4. Let’s start with a problem involving a binomial distribution.

Example 3

The probability of scoring above 75% on a math test is 40%. What is the probability of scoring below 75%?

Solution:

P(scoring above 75%) = 0.40

Therefore, P(scoring below 75%) = 1 - 0.40 = 0.60.

The randomness of an individual outcome occurs when we take 1 event and repeat it over and over again. One example is if you were to flip a coin multiple times. In order to calculate the probability of this type of event, we need to look at one more formula.

The probability of getting x successes in n trials is given by:

P(X = a) = {_n}C_a \times p^a \times q^{(n - a)}

where:

a is the number of successes from the trials.

p is the probability of the event occurring.

q is the probability of the event not occurring.

Now, remember that in Chapter 2, you learned about the formula {_n}C_r. The formula is shown below:

{_n}C_r = \frac{n!}{r!(n-r)!}

Also, recall that the symbol ! means factorial. As a review, the factorial function (!) just means to multiply a series of consecutive descending natural numbers.

Examples:

4! & = 4 \times 3 \times 2 \times 1 = 24\\6! & = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\\1! & = 1

Note: it is generally agreed that 0! = 1.

Technology Tip: You can find the factorial function using:

\boxed{\text{MATH}} \  \boxed{\blacktriangleright}  \ \boxed{\blacktriangleright}  \ \boxed{\blacktriangleright}  \ (\text{PRB}) \boxed{\blacktriangledown}  \ \boxed{\blacktriangledown}  \ \boxed{\blacktriangledown}  \ (\boxed{4})

Now let’s try a few problems with the binomial distribution formula.

Example 4

A fair die is rolled 10 times. Let X be the number of rolls in which we see a 2.

(a) What is the probability of seeing a 2 in any one of the rolls?

(b) What is the probability of seeing a 2 exactly once in the 10 rolls?

Solution:

(a) P(X) = \frac{1}{6} = 0.167

(b) p = 0.167\!\\q =1 - 0.167 = 0.833\!\\n = 10\!\\a = 1

P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(X = 1) & = {_{10}}C_1 \times p^1 \times q^{(10-1)}\\P(X = 1) & = {_{10}}C_1 \times (0.167)^1 \times (0.833)^{(10-1)}\\P(X = 1) & = 10 \times 0.167 \times 0.193\\P(X = 1) & = 0.322

Therefore, the probability of seeing a 2 exactly once when a die is rolled 10 times is 32.2%.

Interestingly, it was Blaise Pascal (pictured below) with Pierre de Fermat who provided the world with the basics of probability. These 2 mathematicians studied many different theories in mathematics, one of which was odds and probability. To learn more about Pascal, go to http://en.wikipedia.org/wiki/Blaise_Pascal. To learn more about Fermat, go to http://en.wikipedia.org/wiki/Fermat. These 2 mathematicians have contributed greatly to the world of mathematics.

Example 5

A fair die is rolled 15 times. What is the probability of rolling two 2’s?

(a) What is the probability of seeing a 2 in any one of the rolls?

(b) What is the probability of seeing a 2 exactly twice in the 15 rolls?

Solution:

(c) P(X) = \frac{1}{6} = 0.167

(d) p = 0.167\!\\q =1 - 0.167 = 0.833\!\\n = 15\!\\a = 2

P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(X = 2) & = {_{15}}C_2 \times p^2 \times q^{(15-2)}\\P(X = 2) & = {_{15}}C_2 \times (0.167)^2 \times (0.833)^{(15-2)}\\P(X = 2) & = 105 \times 0.0279 \times 0.0930\\P(X = 2) & = 0.272

Therefore, the probability of seeing a 2 exactly twice when a die is rolled 15 times is 27.2%.

Example 6

A pair of fair dice is rolled 10 times. Let X be the number of rolls in which we see at least one 2.

(a) What is the probability of seeing at least one 2 in any one roll of the pair of dice?

(b) What is the probability that in exactly half of the 10 rolls, we see at least one 2?

Solution:

If we look at the chart below, we can see the number of times a 2 shows up when rolling 2 dice.

(a) The probability of seeing at least one 2 in any one roll of the pair of dice is:

P(X) = \frac{11}{36} = 0.306

(b) The probability of seeing at least one 2 in exactly 5 of the 10 rolls is calculated as follows:

p & = 0.306\\q & = 1 - 0.306 = 0.694\\n & = 10\\a & = 5

P(X = a) & = {_n}C_a \times p^a \times q^{(n - a)}\\P(X = 5) & = {_{10}}C_5 \times p^5 \times q^{(10-5)}\\P(X = 5) & = {_{10}}C_5 \times (0.306)^5 \times (0.694)^{(10-5)}\\P(X = 5) & = 252 \times 0.00268 \times 0.161\\P(X = 5) & = 0.109

Therefore, the probability of rolling at least one 2 exactly 5 times when 2 dice are rolled 10 times is 10.9%.

It should be noted here that the previous 2 examples are examples of binomial experiments. We will be learning more about binomial experiments and distributions in Chapter 4. For now, we can visualize a binomial distribution experiment as one that has a fixed number of trials, with each trial being independent of the others. In other words, rolling a die twice to see if a 2 appears is a binomial experiment, because there is a fixed number of trials (2), and each roll is independent of the others. Also, for binomial experiments, there are only 2 possible outcomes (a successful event and a non-successful event). For our rolling of the die, a successful event is seeing a 2, and a non-successful event is not seeing a 2.

Example 7

You are given a bag of marbles. Inside the bag are 5 red marbles, 4 white marbles, and 3 blue marbles. Calculate the probability that with 6 trials, you choose 3 marbles that are red, 1 marble that is white, and 2 marbles that are blue, replacing each marble after it is chosen.

Solution:

Notice that this is not a binomial experiment, since there are more than 2 possible outcomes. For binomial experiments, k = 2 (2 outcomes). Therefore, we use the binomial experiment formula for problems involving heads or tails, yes or no, or success or failure. In this problem, there are 3 possible outcomes: red, white, or blue. This type of experiment produces what we call a multinomial distribution. In order to solve this problem, we need to use one more formula:

P & = \frac{n!}{n_1!n_2!n_3!\ldots n_k!} \times \left (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}} \right )

where:

n is the number of trials.

p is the probability for each possible outcome.

k is the number of possible outcomes.

Notice that in this example, k equals 3. If we had only red marbles and white marbles, k would be equal to 2, and we would have a binomial distribution.

The probability of choosing 3 red marbles, 1 white marble, and 2 blue marbles in exactly 6 picks is calculated as follows:

n & = 6 \ (6 \ \text{picks})\\p_1 & = \frac{5}{12} = 0.416 \ (\text{probability of choosing a red marble})\\p_2 & = \frac{4}{12} = 0.333 \ (\text{probability of choosing a white marble})\\p_3 & = \frac{3}{12} = 0.25 \ (\text{probability of choosing a blue marble})\\n_1 & = 3 \ (3 \ \text{red marbles chosen})\\n_2 & = 1 \ (1 \ \text{white marble chosen})\\n_3 & = 2 \ (2 \ \text{blue marbles chosen})\\k & = 3 \ (3 \ \text{possibilities})\\P (x = 6) & = \frac{n!}{n_1!n_2!n_3! \ldots n_k!} \times (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}})\\P(x = 6) & = \frac{6!}{3!1!2!} \times (0.416^3 \times 0.333^1 \times 0.25^2)\\P(x = 6) & = 60 \times 0.0720\times 0.333\times 0.0625\\P(x = 6) & = 0.0899

Therefore, the probability of choosing 3 red marbles, 1 white marble, and 2 blue marbles is 8.99%.

Example 8

You are randomly drawing cards from an ordinary deck of cards. Every time you pick one, you place it back in the deck. You do this 5 times. What is the probability of drawing 1 heart, 1 spade, 1 club, and 2 diamonds?

Solution:

n & = 5 \ (5 \ \text{trials})\\p_1 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a heart})\\p_2 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a spade})\\p_3 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a club})\\p_4 & = \frac{13}{52} = 0.25 \ (\text{probability of drawing a diamond})\\n_1 & = 1 \ (1 \ \text{heart})\\n_2 & = 1 \ (1 \ \text{spade})\\n_3 & = 1 \ (1 \ \text{club})\\n_4 & = 2 \ (2 \ \text{diamonds})\\k & = 1 \ (\text{for each trial, there is only} \ 1 \ \text{possible outcome})\\P (x = 5) & = \frac{n!}{n_1!n_2!n_3! \ldots n_k!} \times (p_1{^{n_1}} \times p_2{^{n_2}} \times p_3{^{n_3}} \ldots p_k{^{n_k}})\\P(x = 5) & = \frac{5!}{1!1!1!2!} \times (0.25^1 \times 0.25^1 \times 0.25^1 \times 0.25^2)\\P(x = 5) & = 60 \times 0.25 \times 0.25\times 0.25\times 0.25\\P(x = 5) & = 0.0586

Therefore, the probability of choosing 1 heart, 1 spade, 1 club, and 2 diamonds is 5.86%.

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