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There are a variety of experiments whose outcomes can be reduced to success or failure. This lesson will explore how real-life situations that satisfy specific conditions can be modeled as binomial experiments. Additionally, methods of determining the probability of a certain number of successes in

n

binomial trials will be presented.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.
Understanding Probability

Theoretical Probability
Experimental Probability
Random Variable

Understanding Descriptive Measures

Expected Value of a Random Variable
Standard Deviation

Understanding Types of Data

Continuous Quantity
Discrete Quantity
Examining Decisions

Other Recommended Readings

Pascals Triangle
Binomial Theorem

Explore

Investigating Distributions Using Experiments

The Galton board is a device patented by Sir Francis Galton. It consists of a set of balls that are dropped from the top of the board. As the balls fall, they move to the left or right every time they bounce off of the pegs embedded in the board until they land in one of the bins at the bottom. Each path option has the same probability.

A Galton Board with a set of marbles ready to be dropped.

Based on the outcomes, try to ask the following questions.

Which bins would be expected to collect the most balls?
About how many balls are expected to land in each bin?
How many paths branch out from each peg?

Discussion

Analyzing the Probabilities of the Values of a Random Variable

A random variable assigns a numerical value to an outcome of a probability experiment. In many situations, it is important to know how likely it is that a random variable will take a specific value. This can be represented by listing or graphing the probability of each value of a random variable. This is called a probability distribution.

Concept

Probability Distribution

A probability distribution of a random variable $X$ is a function that gives the probability of each outcome in the sample space. It can be represented by tables, equations, or graphs. A probability distribution needs to satisfy two conditions to be valid.

The probability of each value of $X$ is greater than or equal to $0$ and less than or equal to $1 .$
For a discrete variable, the sum of the probabilities of all possible values of $X$ is $1 .$

A probability distribution can be either discrete like the geometric distribution and the binomial distribution, or continuous like the normal distribution.

If a probability distribution is based on mathematical models and assumptions, then it is called a theoretical probability distribution. On the other hand, an experimental probability distribution is determined by conducting an experiment.

Example

Consider the roll of a pair of standard dice. Let $X$ be the random variable that represents the sum of the two dice. By the fundamental counting principle, since rolling each die has $6$ possible outcomes, there are a total of $6 \cdot 6 = 36$ possible results. Additionally, the possible values of $X$ are integers from $2$ to $12 .$

A table that represents the theoretical probability distribution of $X$ will now be created. Frequencies represent the number of dice roll results that add up to the given values $x$ of the random variable $X .$ The frequency is divided by $36$ to determine the theoretical probability of each outcome.

$X = Sum of Two Dice$
$x$	Frequency	$P (X = x)$
$2$	$1$	$\frac{1}{3 6} \approx 0.028$
$3$	$2$	$\frac{2}{3 6} \approx 0.056$
$4$	$3$	$\frac{3}{3 6} \approx 0.083$
$5$	$4$	$\frac{4}{3 6} \approx 0.111$
$6$	$5$	$\frac{5}{3 6} \approx 0.139$
$7$	$6$	$\frac{6}{3 6} \approx 0.167$
$8$	$5$	$\frac{5}{3 6} \approx 0.139$
$9$	$4$	$\frac{4}{3 6} \approx 0.111$
$10$	$3$	$\frac{3}{3 6} \approx 0.083$
$11$	$2$	$\frac{2}{3 6} \approx 0.056$
$12$	$1$	$\frac{1}{3 6} \approx 0.028$

The probability distribution of

X

can also be represented by a bar graph. The possible outcomes of

X

are marked on the horizontal axis and their probabilities are presented on the vertical axis.

In this example, each possible value of the random variable can be associated with its corresponding probability because it is a discrete random variable. This is also why the bars of the probability distribution must be separated.

Example

Determine the Theoretical and Experimental Probability Distributions of a Coin Toss

Izabella is a big soccer fan. One weekend, she invited her friend Dylan to watch a world championship match together at her house. During the coin toss ceremony, Izabella asked Dylan about the number of heads they will obtain if they toss a fair coin four times.

Let $X$ be a random variable that represents the number of heads in four coin flips. Help Izabella and Dylan solve the following problems and determine whether they can predict the number of times the experiment results in heads.

a Construct a table to describe the theoretical probability distribution of

X .

b Graph the theoretical probability distribution of

X .

c Izabella and Dylan conducted an experiment of tossing the fair coin four times. They repeated this experiment

100

times and recorded the data in a tally sheet.

Number of Heads, $x$	Tally	Frequency
$0$	$∣ ∣ ∣ ∣ ∣$	$6$
$1$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$22$
$2$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$37$
$3$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$28$
$4$	$∣ ∣ ∣ ∣ ∣ ∣$	$7$

Use this data to find the experimental probability of each possible value of $X .$

d Graph the experimental probability distribution of

X .

Answer

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.0625$	$0.25$	$0.375$	$0.25$	$0.0625$

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.06$	$0.22$	$0.37$	$0.28$	$0.07$

Hint

a Think about the possible outcomes of tossing a coin four times. How can the outcomes be represented?

b In a discrete distribution, the bars on the graph will be separated.

c Divide the frequency of each outcome by the total number of trials.

d The horizontal axis will represent the possible outcomes and the vertical axis will represent the probability of each outcome.

Solution

a The theoretical distribution of

X

will be determined by using a table. Observe the following simulation and think about the possible values for

X .

Keep in mind that there are exactly four coin flips.

Because

X

represents the number of heads in four coin flips, the possible values of the variable are

0,

1,

2,

3,

and

4 .

To analyze the outcomes where each value is obtained, mark heads as

H

and tails as

T .

Then, write all possible outcomes for each number of heads and count them.

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
Possible Outcomes	$T T T T$	$T T T H,$ $T T H T,$ $T H T T,$ $H T T T$	$H H T T,$ $H T T H,$ $T T H H,$ $H T H T,$ $T H T H,$ $T H H T$	$H H H T,$ $H H T H,$ $H T H H,$ $T H H H$	$H H H H$
Frequency	$1$	$4$	$6$	$4$	$1$

Before calculating the theoretical probability of each value, find the total number of possible outcomes.

Total : 1 + 4 + 6 + 4 + 1 = 16

Now divide the frequency of each outcome by the total number of possibilities to calculate the probability

P (X = x)

that the random variable

X

takes the specific value

x .

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
Frequency	$1$	$4$	$6$	$4$	$1$
$P (X = x)$	$\frac{1}{1 6}$	$\frac{4}{1 6} = \frac{1}{4}$	$\frac{6}{1 6} = \frac{3}{8}$	$\frac{4}{1 6} = \frac{1}{4}$	$\frac{1}{1 6}$

This table describes the theoretical probabilities associated with tossing a fair coin four times. Since only the theoretical probability is required, the Frequency column can be skipped.

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.0625$	$0.25$	$0.375$	$0.25$	$0.0625$

b To graph the distribution, each possible outcome will be represented on the horizontal axis and the probability of each outcome on the vertical axis. Since only

0,

1,

2,

3,

and

4

instances of heads are possible — it cannot appear

1.5

times out of

4

tosses, for example — the probability distribution is discrete and the bars on the graph will be separated.

c Consider the tally sheet prepared by Izabella and Dylan when conducting

100

trials of the experiment consisting of tossing a fair coin four times.

Number of Heads, $x$	Tally	Frequency
$0$	$∣ ∣ ∣ ∣ ∣$	$6$
$1$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$22$
$2$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$37$
$3$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$28$
$4$	$∣ ∣ ∣ ∣ ∣ ∣$	$7$

Calculate the experimental probability of each possible outcome by dividing its frequency by the total number of trials, $100 .$

Number of Heads, $x$	Tally	Frequency	$P (X = x)$
$0$	$∣ ∣ ∣ ∣ ∣$	$6$	$\frac{6}{1 0 0}$
$1$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$22$	$\frac{2 2}{1 0 0}$
$2$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$37$	$\frac{3 7}{1 0 0}$
$3$	$∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$	$28$	$\frac{2 8}{1 0 0}$
$4$	$∣ ∣ ∣ ∣ ∣ ∣$	$7$	$\frac{7}{1 0 0}$

Since only the experimental probability is required, the Tally and Frequency columns can be skipped. Next, write the table horizontally.

	$X = Number of Heads$
$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$0.06$	$0.22$	$0.37$	$0.28$	$0.07$

d Follow a similar method as in Part B to graph the probability distribution. Again, the horizontal axis will represent the possible values of

X,

while the vertical axis will represent the probability of each outcome.

From the graph, it can be seen that the outcome with the highest probability of occurring is

2 .

This implies that

2

will be the most common outcome as the experiment is continued to be performed. This matches what the theoretical probability calculated previously.

Discussion

Can the Experimental Distribution Estimate the Theoretical Distribution?

The expected value of a random variable $X$ is the average of the possible outcomes of a random variable. It is used to describe the center of a probability distribution. For a discrete random variable, the expected value $E (X)$ is given by the weighted mean.

$E (X) = i = 1 \sum n x_{i} \cdot P (X = x_{i})$

In this formula, $x_{i}$ represents a specific outcome, $P (X = x_{i})$ corresponds to the associated probability of $x_{i},$ and $n$ is the number of all possible outcomes. According to the law of large numbers, when considering a sequence of random variables, its average tends to the expected value under specific conditions.

Rule

Law of Large Numbers

Let

X_{1},

X_{2},

\dots,

X_{n}

be independent random variables with an identical probability distribution and

\overline{S_{n}}

be the mean of the random variables.

\overline{S_{n}} = \frac{X _{1} + X _{2} + \dots + X _{n}}{n}

The independence of random variables means that the events that different random variables take particular values are independent. If

E (X_{i}) = μ

represents the expected value of

X_{i},

the law of large numbers states that

\overline{S_{n}}

tends towards

μ .

$n \to + \infty lim \overline{S_{n}} = μ$

This notation means that for a sufficiently large number of observations

n

— or by repeating the experiment a large number of times — the mean

\overline{S_{n}}

will approximate the expected value

μ .

This law can be used to find the theoretical probabilities of the outcomes of an experiment by using the experimental probabilities. Consider the flipping of a fair coin experiment.

Suppose it is unknown that the probability of landing on either heads or tails on a single trial is

0.5 .

For a few coin tosses, the experimental probability might not come anywhere near

0.5 .

An applet showing how the experimental probabilities of the possible outcomes vary after each trial.

However, as the number of trials increases, the proportion of heads and tails outcomes will be closer to their theoretical probabilities. Suppose that this experiment is repeated

100,

500,

or even

1000

times, then consider the event that the outcome of the coin toss is tails.

An applet showing how the experimental probability of tails gets closer to its theoretical probability as the number of trials increases.

This law helps understand why casinos always win. Someone can be lucky and win a certain amount of times, but in the long run, the casino's earnings will converge to a predictable percentage, the expected value. The house always wins.

Discussion

How is the Variation of a Probability Distribution Measured?

The expected value is commonly used with a measure of variation such as the variance or standard deviation to determine how outcome will differ from the expected value.

Concept

Variance of a Random Variable

The variance of a random variable describes how far from the expected value

E (X)

the outcomes of a random variable

X

are likely to be. To calculate the variance, begin by determining the deviation of each possible outcome

x_{i}

— the difference between

x_{i}

and

E (X) .

\underline{Deviation of x_{i}} x_{i} - E (X)

The variance is the total sum of the products of the deviation of each outcome and its corresponding probability

P (X = x_{i}) .

σ^{2} = Σ [[x_{i} - E (X)]^{2} \cdot P (X = x_{i})]

The variance is denoted by

σ^{2}

because it is the square of the standard deviation

σ .

Concept

Standard Deviation of a Random Variable

The standard deviation of a random variable is a measure of variation that describes how spread out the outcomes of a random variable $X$ are from its expected value $E (X) .$ The standard deviation is represented by the Greek letter $σ$ — read as sigma — and is given by the square root of the variance of $X .$

σ = Σ [[x_{i} - E (X)]^{2} \cdot P (X = x_{i})]

In this formula, $x_{i}$ is a specific outcome and $P (X = x_{i})$ is the probability of $x_{i} .$

Example

Let $X$ be the random variable representing the number of cars sold on a given day in a car dealership. The table below shows the probability distribution of $X .$

$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$\frac{1}{1 5}$	$\frac{3}{1 5}$	$\frac{6}{1 5}$	$\frac{3}{1 5}$	$\frac{2}{1 5}$

Use the probability distribution to calculate the standard deviation of

X .

Begin by finding the expected value.

E (X) = [x_{1} \cdot P (X = x_{1})] + [x_{2} \cdot P (X = x_{2})] + \dots + [x_{n} \cdot P (X = x_{n})]

SubstituteValues

Substitute values

E (X) = 0 (\frac{1}{1 5}) + 1 (\frac{3}{1 5}) + 2 (\frac{6}{1 5}) + 3 (\frac{3}{1 5}) + 4 (\frac{2}{1 5})

▼

Evaluate right-hand side

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

E (X) = \frac{0}{1 5} + \frac{3}{1 5} + \frac{1 2}{1 5} + \frac{9}{1 5} + \frac{8}{1 5}

$\frac{0}{a} = 0$

E (X) = 0 + \frac{3}{1 5} + \frac{1 2}{1 5} + \frac{9}{1 5} + \frac{8}{1 5}

IdPropAdd

Identity Property of Addition

E (X) = \frac{3}{1 5} + \frac{1 2}{1 5} + \frac{9}{1 5} + \frac{8}{1 5}

AddFrac

Add fractions

E (X) = \frac{3 2}{1 5}

UseCalc

Use a calculator

E (X) = 2.1333 \dots

RoundDec

Round to $2$ decimal place(s)

E (X) \approx 2.13

Next, the variance of

X

will be calculated using a table of values. To do so, calculate the square of each deviation — the difference between each possible value of

X

and

E (X) .

Then multiply the square deviation of each value by its corresponding probability.

$x_{i}$	$[x_{i} - E (X)]^{2}$	$[x_{i} - E (X)]^{2} \cdot P (X = x_{i})$
$0$	$(0 - 2.13)^{2} = 4.5369$	$4.5364 \cdot \frac{1}{1 5} \approx 0.3025$
$1$	$(1 - 2.13)^{2} = 1.2769$	$1.2769 \cdot \frac{3}{1 5} \approx 0.2554$
$2$	$(2 - 2.13)^{2} = 0.0169$	$0.0169 \cdot \frac{6}{1 5} \approx 0.0068$
$3$	$(3 - 2.13)^{2} = 0.7569$	$0.7569 \cdot \frac{3}{1 5} \approx 0.1514$
$4$	$(4 - 2.13)^{2} = 3.4969$	$3.4969 \cdot \frac{2}{1 5} \approx 0.4663$
Variance $σ^{2}$		$\approx 1.1824$

Finally, calculate the square root of the variance to get the standard deviation of $X .$

\underline{Standard Deviation :} 1.1824 \approx 1.09

Variance vs Standard Deviation

The standard deviation is preferred over the variance because taking the square root of the variance gives the standard deviation the same units as

X .

This makes it possible to compare the possible outcomes relative to the expected value.

Example

Using Standard Deviation to Make a Decision

Izabella's aunt Magdalena owns a clothing store. She needs to increase stock in her shop and plans to invest $$ 15000$ in one of the two collections that were offered to her by well-known brands. Each brand claims that they have a great expected rate of return. Their probability distributions are described below.

Dylan and Izabella want to help Magdalena make the best decision. They decided to use their recently acquired knowledge about the expected value and standard deviation of a probability distribution to analyze the offers. Help them answer the following questions and give the best advice to Magdalena.

a Pair each description with its corresponding measure.

b Which investment should Izabella and Dylan advise Magdalena to choose?

Hint

a The expected value is the total sum of the products of every possible value of the random variable and its corresponding probability.

b Compare the expected values and the standard deviations of the distributions.

Solution

a To match each description with its corresponding value, find the expected value and the standard deviation of each probability distribution one at a time.

Collection I

The expected value

E (X)

of a discrete random variable is given by the total sum of the products of every possible value

x_{i}

and its associated probability

P (X = x_{i}) .

Let

X

represent the profit of the first collection. Note that each loss will be represented by a negative value.

E (X) = [x_{1} \cdot P (X = x_{1})] + [x_{2} \cdot P (X = x_{2})] + \dots + [x_{n} \cdot P (X = x_{n})]

SubstituteValues

Substitute values

E (X) = 1200 \cdot 0.5 + 1800 \cdot 0.2 + 900 \cdot 0.2 + (- 150) \cdot 0.1

Multiply

E (X) = 600 + 360 + 180 - 15

AddSubTerms

Add and subtract terms

E (X) = 1125

The expected value can now be used to find the standard deviation of the probability distribution. Consider the formula for the standard deviation.

σ = Σ [[x_{i} - E (X)]^{2} \cdot P (X = x_{i})]

To apply this formula, calculate the square of each deviation — the difference between each

x_{i}

value and the expected value — and then multiply it by its corresponding probability. This process will be shown in a table.

$x_{i}$	$[x_{i} - E (X)]^{2}$	$[x_{i} - E (X)]^{2} \cdot P (X = x_{i})$
$1200$	$(1200 - 1125)^{2} = 5625$	$5625 \cdot 0.5 = 2812.5$
$1800$	$(1800 - 1125)^{2} = 455625$	$455625 \cdot 0.2 = 91125$
$900$	$(900 - 1125)^{2} = 50625$	$50625 \cdot 0.2 = 10125$
$- 150$	$(- 150 - 1125)^{2} = 1625625$	$1625625 \cdot 0.1 = 162562.5$
Sum of Values		$266625$

The sum of the values of the last column represents the variance of the probability distribution. The standard deviation will be found by taking the square root of

266625 .

\underline{Standard Deviation :} 266625 \approx 516.36

Collection II

Follow a similar procedure as before to calculate the expected value for Collection II.

E (X) = [x_{1} \cdot P (X = x_{1})] + [x_{2} \cdot P (X = x_{2})] + \dots + [x_{n} \cdot P (X = x_{n})]

SubstituteValues

Substitute values

E (X) = 3600 \cdot 0.3 + 2850 \cdot 0.1 + (- 300) \cdot 0.4 + (- 500) \cdot 0.2

Multiply

E (X) = 1080 + 285 - 120 - 100

AddSubTerms

Add and subtract terms

E (X) = 1145

Now use a table to calculate the variance of the distribution.

$x_{i}$	$[x_{i} - E (X)]^{2}$	$[x_{i} - E (X)]^{2} \cdot P (X = x_{i})$
$3600$	$(3600 - 1145)^{2} = 6027025$	$6027025 \cdot 0.3 = 1808107.5$
$2850$	$(2850 - 1145)^{2} = 2907025$	$2907025 \cdot 0.1 = 290702.5$
$- 300$	$(- 300 - 1145)^{2} = 2088025$	$2088025 \cdot 0.4 = 835210$
$- 500$	$(- 500 - 1145)^{2} = 2706025$	$2706025 \cdot 0.2 = 541205$
Sum of Values		$3475225$

The square root of the variance will be calculated to find the standard deviation of the probability distribution of Collection II.

\underline{Standard Deviation :} 3475225 \approx 1864.20

Conclusion

The expected value and the standard distribution of each probability distribution have been calculated. The following table summarizes these measures.

Measures of the Probability Distributions
Expected Value of Collection I	$1125$
Expected Value of Collection II	$1145$
Standard Deviation of Collection I	$\approx 516.23$
Standard Deviation of Collection II	$\approx 1864.20$

b Consider the expected value of each distribution.

\underline{Expected Values} Collection I : Collection II : E (X) = 1125 E (X) = 1145

The expected values do not give much information on their own because they are very close. This means that the expected profit of each collection is similar. Next, compare the standard deviations to determine which distribution has more variability.

\underline{Standard Deviation} Collection I : Collection II : σ \approx 516.36 σ \approx 1864.20

The standard deviation of Collection II is almost four times the that of Collection I, which implies that the expected value of Collection II will have about four times the variability of Collection I. Since Collection II is riskier, with a high chance for both gains and losses, Izabella and Dylan should advise Magdalena to invest in Collection I.

Discussion

Binomial Experiments

The outcomes of many experiments can be reduced to two possibilities, success or failure. If two more conditions are satisfied, these experiments can be modeled by a binomial experiment.

Concept

Binomial Experiment

A binomial experiment is a probability experiment that has the following three properties.

There is a fixed number of independent trials.
Each trial has exactly two possible outcomes — success and failure.
For each trial, the probability of success is constant.

Details of each property can be seen in the applet by selecting the corresponding number.

The number of successes in a binomial experiment is a random variable that has a binomial distribution.

Example

One of the most well-known examples of a binomial experiment is flipping a coin. There are only two possible outcomes, heads and tales. Either heads or tails could be considered a success, depending on the situation. The outcome of one toss does not impact the probability of the outcome of the next trial.

Extra

An Experiment Reducible to a Binomial Experiment

Note that many probability experiments can be reduced so that they satisfy the conditions of a binomial experiment. In case of rolling a die, there are six possible outcomes. However, they can be divided into two groups — even numbers and odd numbers, for example.

After grouping the outcomes, there are two possible results of each trial — rolling an even number or an odd number. The probability of rolling a number from either group is constant throughout the trials, and the result of rolling the die is not affected by the previous results.

Therefore, rolling a die can also be considered as a binomial experiment as long as success and failure are clearly defined.

Example

Classifying Whether Experiments Are Binomial

To better understand binomial experiments, Dylan and Izabella listed some examples of experiments. They want to determine whether the experiments can be modeled as binomial experiments. Which of these situations are examples of binomial experiments?

Hint

Is there a fixed number of trials for each experiment? How many outcomes are possible? Does the probability of each outcome remain constant for each trial? Are trials independent?

Solution

Start by recalling the conditions that a binomial experiment should satisfy.

There is a fixed number of independent trials.
Each trial has exactly two possible outcomes — success and failure.
For each trial, the probability of success is constant.

Analyze each situation one at a time to see if it follows all of the conditions of a binomial experiment.

The Scratch-off Cards

This situation has eight trials of selecting one scratch-off cards at random.

Each card could win a prize or not, which means there are two possible outcomes for each trial. Moreover, the probability of success, which is winning a prize, is

25 %

0.25,

for every card.

P (Success) = 0.25

Finally, the trials are independent because scratching one card does not affect the probability that any of the other cards reveals a prize. Therefore, this situation represents a binomial experiment.

Heights of $100$ People

Note that height can vary for every people surveyed.

Because there are $100$ possible answers, it is likely that more than $2$ different outcomes will occur. This means that this situation does not represent a binomial experiment.

Rolling a Die

In this case, rolling a

3

can take only one or many more trials.

It cannot be known how many rolls it will take until a

3

comes up. Therefore the number of trials is not fixed. This means that this experiment does not represent a binomial experiment.

Blood Type O

This situation has a fixed number of trials because it involves asking $20$ people if their blood type is O.

Each trial has only two possible solutions, blood type O or another type. Moreover, the probability of having blood type O in each trial is

0.4,

which represents the probability of success.

P (Success) = 0.4

Finally, because the answer of any person surveyed does not affect the probability that other people have O type blood, each trial is independent. This means that this situation is a binomial experiment.

Conclusion

All four situations have already been analyzed and the results are summarized in the following table.

Experiment	Is It a Binomial Experiment?
The Scratch-off Cards	$✓$
Heights of $100$ People	$\times$
Rolling a Die	$\times$
Blood Type O?	$✓$

Discussion

Probability of $x$ Successes in $n$ Binomial Trials

Because binomial experiments can simplify many complex situations, it is essential to determine how likely it is to obtain a specific number of successes out of $n$ trials in a given experiment. Also, the expected value, or center of the distribution, will be presented.

Concept

Binomial Distribution

The binomial distribution is the probability distribution that describes the number of successes $x$ out of $n$ binomial trials. The trials must satisfy three conditions.

There is a fixed number of independent trials.
Each trial has exactly two possible outcomes — success and failure.
The probability of success is constant for each trial.

Let $X$ be a random variable representing the total number of successes among $n$ trials. The possible values of $X$ are $x = 0,$ $1,$ $2,$ $\dots,$ $n .$ The binomial probability formula can be used to determine the probability of $x$ successes among $n$ trials $P (X = x) .$

$P (X = x) =_{n} C_{x} p^{x} q^{n - x}$

In this formula, $_{n} C_{x}$ is the binomial coefficient and $p$ and $q$ are the probabilities of success and failure, respectively. Additionally, the expected value of $X$ can be determined by the product of the number of trials $n$ and the probability of success $p .$

$E (X) = n p$

This means that the expected number of successes in $n$ trials is given by $n p .$

Example

Consider the experiment of drawing $1$ card from a standard deck of cards with replacement.

If drawing a diamond is considered a success, let

X

be the number of diamonds drawn. Since there are

13

diamond cards in a standard deck, the probability of success

p

in each trial will be

\frac{1 3}{5 2} = \frac{1}{4} .

The probability of failure

q

will be

1 - \frac{1}{4} = \frac{3}{4} .

Suppose the experiment is repeated

5

times. Then, the binomial distribution with

n = 5

can be determined.

P (X = x) =_{5} C_{x} (\frac{1}{4})^{x} (\frac{3}{4})^{5 - x}

In this situation, the possible values of

x

are

0,

1,

2,

3,

4,

and

5 .

The distribution of the random variable

X

can be obtained by evaluating this formula for each of these values.

$x$	$_{5} C_{x} (\frac{1}{4})^{x} (\frac{3}{4})^{5 - x}$	$P (X = x) =_{5} C_{x} (\frac{1}{4})^{x} (\frac{3}{4})^{5 - x}$
$0$	$_{5} C_{0} (\frac{1}{4})^{0} (\frac{3}{4})^{5 - 0} = \frac{2 4 3}{1 0 2 4}$	$\approx 0.237$
$1$	$_{5} C_{1} (\frac{1}{4})^{1} (\frac{3}{4})^{5 - 1} = \frac{4 0 5}{1 0 2 4}$	$\approx 0.396$
$2$	$_{5} C_{2} (\frac{1}{4})^{2} (\frac{3}{4})^{5 - 2} = \frac{2 7 0}{1 0 2 4}$	$\approx 0.264$
$3$	$_{5} C_{3} (\frac{1}{4})^{3} (\frac{3}{4})^{5 - 3} = \frac{9 0}{1 0 2 4}$	$\approx 0.088$
$4$	$_{5} C_{4} (\frac{1}{4})^{4} (\frac{3}{4})^{5 - 4} = \frac{1 5}{1 0 2 4}$	$\approx 0.015$
$5$	$_{5} C_{5} (\frac{1}{4})^{5} (\frac{3}{4})^{5 - 5} = \frac{1}{1 0 2 4}$	$\approx 0.001$

The following graph shows the probability distribution of

X .

Binomial Distribution of Diamonds in 10 Drawns with Replacement

Extra

Deriving the Expected Value of a Binomial Distribution

The expected value

E (X)

of a discrete random variable

X

is given by the sum of the products of every possible value

x_{i}

of the variable and its corresponding probability

P (X = x_{i}) .

E (X) = i = 1 \sum n x_{i} \cdot P (X = x_{i})

Because the number of successes out of

n

trials can be

x = 0,

1,

\dots,

n

in a binomial experiment, the expected value of the binomial random variable can be simplified as follows.

E (X) = x = 0 \sum n x \cdot P (X = x)

The binomial probability formula gives that

P (X = x)

is equal to

_{n} C_{x} p^{x} q^{n - x} .

This expression can be substituted into the simplified form of the expected value of a binomial distribution.

E (X) = x = 0 \sum n x \cdot P (X = x) ⇓ E (X) = x = 0 \sum n x \cdot_{n} C_{x} p^{x} q^{n - x}

Note that when

x

is equal to

0,

the expression

x \cdot_{n} C_{x} p^{x_{i}} q^{n - x} = 0 .

This means that the first addend of the sum can be omitted and the first term of the sum is calculated for

x = 1 .

E (X) = x = 1 \sum n x \cdot_{n} C_{x} p^{x} q^{n - x}

This formula can be manipulated to find the expected value of the binomial distribution.

E (X) = x = 1 \sum n x \cdot_{n} C_{x} \cdot p^{x} q^{n - x}

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

E (X) = x = 1 \sum n x \cdot \frac{n !}{x ! ( n - x ) !} \cdot p^{x} q^{n - x}

▼

Rewrite

Write as a product

E (X) = x = 1 \sum n x \cdot \frac{n \cdot ( n - 1 ) !}{x \cdot ( x - 1 ) ! ( n - x ) !} \cdot p^{x} q^{n - x}

CancelCommonFac

Cancel out common factors

E (X) = x = 1 \sum n x \cdot \frac{n \cdot ( n - 1 ) !}{x \cdot ( x - 1 ) ! ( n - x ) !} \cdot p^{x} q^{n - x}

SimpQuot

Simplify quotient

E (X) = x = 1 \sum n \frac{n \cdot ( n - 1 ) !}{( x - 1 ) ! ( n - x ) !} \cdot p^{x} q^{n - x}

Rewrite

Rewrite $x$ as $(x - 1) + 1$

E (X) = x = 1 \sum n \frac{n \cdot ( n - 1 ) !}{( x - 1 ) ! ( n - x ) !} \cdot p^{(x - 1) + 1} q^{n - x}

SumInExponentOne

$a^{1 + m} = a \cdot a^{m}$

E (X) = x = 1 \sum n \frac{n \cdot ( n - 1 ) !}{( x - 1 ) ! ( n - x ) !} \cdot p \cdot p^{x - 1} q^{n - x}

MovePartNumLeft

$\frac{a \cdot b}{c} = a \cdot \frac{b}{c}$

E (X) = x = 1 \sum n n \cdot \frac{( n - 1 ) !}{( x - 1 ) ! ( n - x ) !} \cdot p \cdot p^{x - 1} q^{n - x}

CommutativePropMult

Commutative Property of Multiplication

E (X) = x = 1 \sum n n p \cdot \frac{( n - 1 ) !}{( x - 1 ) ! ( n - x ) !} \cdot p^{x - 1} q^{n - x}

FactorOut

Factor out $n p$

E (X) = n p x = 1 \sum n (\frac{( n - 1 ) !}{( x - 1 ) ! ( n - x ) !}) p^{x - 1} q^{n - x}

Rewrite

Rewrite $n - x$ as $n - 1 - (x - 1)$

E (X) = n p x = 1 \sum n (\frac{( n - 1 ) !}{( x - 1 ) ! ( ( n - 1 ) - ( x - 1 ) ) !}) p^{x - 1} q^{(n - 1) - (x - 1)}

Now, substitute

y = x - 1

and

m = n - 1 .

If x = 1, then y = 0 If x = n, then y = n - 1 = m

The sum in the last expression can now be rewritten in terms of

y

and

m

by using this information.

E (X) = n p x = 1 \sum n (\frac{( n - 1 ) !}{( x - 1 ) ! ( ( n - 1 ) - ( x - 1 ) ) !}) p^{x - 1} q^{(n - 1) - (x - 1)}

SubstituteValues

Substitute values

E (X) = n p y = 0 \sum m (\frac{m !}{y ! ( n - y ) !}) p^{y} q^{n - y}

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

E (X) = n p y = 0 \sum m_{m} C_{y} p^{y} q^{n - y}

Note that the sum corresponds to the Binomial Theorem.

Next, recall that

p + q = 1

and that the powers of

1

are all equal to

1 .

Therefore, the sum will equal

1 .

y = 0 \sum m_{m} C_{y} p^{y} q^{n - y} y = 0 \sum m_{m} C_{y} p^{y} q^{n - y} = (p + q)^{m} ⇓ = 1

Substituting this into the expression of the expected value results in the formula for the expected value of a binomial distribution.

E (X) = n p y = 0 \sum m_{m} C_{y} p^{y} q^{n - y}

Substitute

$y = 0 \sum m_{m} C_{y} p^{y} q^{n - y} = 1$

E (X) = n p \cdot 1

MultByOne

$a \cdot 1 = a$

E (X) = n p

Rule

Binomial Probability

Let $X$ be the random variable representing the number of successes in $n$ binomial trials. The binomial probability $P (X = x)$ can be calculated by using the following formula.

$P (X = x) =_{n} C_{x} p^{x} q^{n - x}$

In this formula, $P (X = x)$ is the probability that the random variable $X$ is equal to $x,$ which means that there are exactly $x$ successes. Additionally, $p$ and $q$ are the probabilities of success and failure, respectively, and $_{n} C_{x}$ is the binomial coefficient.

Proof

This proof will begin by calculating the probability of obtaining a fixed sequence with $x$ successes in $n$ independent trials. The number of all possible sequences with $x$ successes will then be calculated. Finally, joining the first and second parts will give the formula.

The Probability for a Fixed Sequence

Note that there are

n

independent trials, of which

x

are successes. This means that the difference between

n

and

x

gives the number of failures.

Number of Succeses : Number of Failures : x n - x

Furthermore, let

p

be the probability of success and

q

be the probability of failure in one individual trial. Therefore, by the Multiplication Rule of Probability and the fact that trials are independent, the probability of

x

successes is given by multiplying

p

by itself

x

times.

Probability of x Successes : x times p \cdot p \cdot \dots \cdot p = p^{x}

Similarly, because there are

n - x

failures, the probability of

n - x

failures is given by multiplying

q

by itself

n - x

times.

Probability of n - x Failures : n - x times q \cdot q \cdot \dots \cdot q = q^{n - x}

Therefore, the probability of getting exactly

x

successes and

n - x

failures is given by the product of

p^{x}

and

q^{n - x}

p^{x} q^{n - x}

By the Commutative Property of Multiplication, any combination of

p

and

q

variables can be rearranged. Therefore, the expression is valid for any fixed sequence of

x

successes and

n - x

failures. However, note that this is the probability of only one of the possible sequences.

Finding the Total Number of Possible Sequences

Consider an experiment that consists of four independent trials. The sequences in which two successes

S

and two failures

F

occur are as follows.

{S S F F, S F S F, S F F S, - F S S F, F S F S, F F S S}

Note that as long as two successes occur, the order of the outcomes is not important. This can be simplified by considering how many ways the two successes can be arranged within the four trials since the remaining trials will automatically be failures. In such a situation, the combination formula can be used.

_{n} C_{r} = \frac{n !}{r ! ( n - r ) !}

SubstituteII

$n = 4$ , $r = 2$

_{4} C_{2} = \frac{4 !}{2 ! ( 4 - 2 ) !}

▼

Evaluate right-hand side

SubTerm

Subtract term

_{4} C_{2} = \frac{4 !}{2 ! \cdot 2 !}

Write as a product

_{4} C_{2} = \frac{4 \cdot 3 \cdot 2 !}{2 ! \cdot 2 !}

CrossCommonFac

Cross out common factors

_{4} C_{2} = \frac{4 \cdot 3 \cdot 2 !}{2 ! \cdot 2 !}

CancelCommonFac

Cancel out common factors

_{4} C_{2} = \frac{4 \cdot 3}{2 !}

Multiply

_{4} C_{2} = \frac{1 2}{2 !}

$2! = 2$

_{4} C_{2} = \frac{1 2}{2}

CalcQuot

Calculate quotient

_{4} C_{2} = 6

This means that

6

different arrangements are possible, which corresponds to the number of sequences listed before. By following the same reasoning, different sequences of

n

trials with

x

successes can be found by calculating the number of combinations of

x

out of

n .

_{n} C_{x} = \frac{n !}{x ! ( n - x ) !}

Getting the Binomial Probability

The probability of obtaining a fixed sequence of

x

successes out of

n

trials is given by the product of

p

to the power of

x

and

q

to the power of

n - x .

p^{x} q^{n - x}

Moreover, there are

_{n} C_{x}

possible sequences with

x

successes. Therefore, by the Addition Rule of Probability, the probability

P (X = x)

of getting any of the possible sequences is given by adding the probability of getting one sequence

_{n} C_{x}

times.

P (X = x) =_{n} C_{x} times p^{x} q^{n - x} + \dots + p^{x} q^{n - x} ⇓ P (X = x) =_{n} C_{x} p^{x} q^{n - x}

Example

Finding the Probability of Guessing Correctly on a Multiple-Choice Test

Dylan is taking a test that consists of five multiple-choice questions. Each question has five answer choices, only one of which is correct. He did not have time to study, so he decides to guess every answer.

A paper with 5 questions and each question has 5 choices

What is the probability of guessing at least three answers correctly? Write the answer as a percentage. Round to two decimal places.

Hint

Consider using the Addition Rule of Probability.

Solution

To find the probability of guessing at least three answers correctly in a five question multiple-choice test, first, verify if this experiment satisfies the three conditions of a binomial experiment.

Condition	Given Experiment	Is Satisfied?
There is a fixed number of independent trials.	There are five trials — five questions.	$✓$
Each trial has two possible outcomes.	The answer for each question is either correct or incorrect.	$✓$
The probability of success is constant for each trial.	The probability of guessing the correct answer for each question is $\frac{1}{5} = 0.2,$ where $5$ is the number of options.	$✓$

Let

X

be a random variable representing the number of correct guesses out of the

5

trials. The possible values of

X

are

0,

1,

2,

3,

4,

and

5 .

Additionally, the probability of failure

q

will be

1 - 0.2 = 0.8 .

Substitute this information into the Binomial Probability Formula.

P (X = x) =_{n} C_{x} p^{x} q^{n - x} ⇓ P (X = x) =_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}

The probability of guessing at least three questions correctly can be written as

P (X \geq 3) .

According to the Addition Rule of Probability, the sum of

P (X = 3),

P (X = 4),

and

P (X = 5)

is equal to the probability of

P (X \geq 3),

as there are exactly

5

questions. Begin by calculating

P (X = 3) .

P (X = x) =_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}

Substitute

$x = 3$

P (X = 3) =_{5} C_{3} (0.2)^{3} (0.8)^{5 - 3}

▼

Simplify right-hand side

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

P (X = 3) = (\frac{5 !}{3 ! ( 5 - 3 ) !}) (0.2)^{3} (0.8)^{5 - 3}

SubTerms

Subtract terms

P (X = 3) = (\frac{5 !}{3 ! \cdot 2 !}) (0.2)^{3} (0.8)^{2}

Write as a product

P (X = 3) = (\frac{5 \cdot 4 \cdot 3 !}{3 ! \cdot 2 \cdot 1 !}) (0.2)^{3} (0.8)^{2}

CancelCommonFac

Cancel out common factors

P (X = 3) = (\frac{5 \cdot 4 \cdot 3 !}{3 ! \cdot 2 \cdot 1 !}) (0.2)^{3} (0.8)^{2}

SimpQuot

Simplify quotient

P (X = 3) = (\frac{5 \cdot 4}{2 \cdot 1 !}) (0.2)^{3} (0.8)^{2}

▼

Evaluate right-hand side

$1! = 1$

P (X = 3) = (\frac{5 \cdot 4}{2 \cdot 1}) (0.2)^{3} (0.8)^{2}

Multiply

P (X = 3) = (\frac{2 0}{2}) (0.2)^{3} (0.8)^{2}

CalcQuot

Calculate quotient

P (X = 3) = (10) (0.2)^{3} (0.8)^{2}

CalcPowProd

Calculate power and product

P (X = 3) = 0.0512

Follow a similar process to find

P (X = 4)

and

P (X = 5) .

$x$	$_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}$	$P (X = x) =_{5} C_{x} (0.2)^{x} (0.8)^{5 - x}$
$3$	$_{5} C_{3} (0.2)^{3} (0.8)^{5 - 3} = \frac{1 6 0}{3 1 2 5}$	$P (X = 3) = 0.0512$
$4$	$_{5} C_{4} (0.2)^{4} (0.8)^{5 - 4} = \frac{2 0}{3 1 2 5}$	$P (X = 4) = 0.0064$
$5$	$_{5} C_{5} (0.2)^{5} (0.8)^{5 - 5} = \frac{1}{3 1 2 5}$	$P (X = 5) \approx 0.00032$

Now, add the probabilities to get

P (X \geq 3) .

P (X \geq 3) = P (X = 3) + P (X = 4) + P (X = 5)

SubstituteValues

Substitute values

P (X \geq 3) \approx 0.0512 + 0.0064 + 0.00032

AddTerms

Add terms

P (X \geq 3) \approx 0.05792

WritePercent

Convert to percent

P (X \geq 3) \approx 5.79 %

Therefore, the probability of Dylan guessing at least

3

correct answers is about

5.79 % .

Example

Finding the Number of Customers That Will Purchase Something

Izabella and Dylan are doing a fantastic job helping out at the clothing store. They noticed that the behavior of the customers can be modeled by a binomial experiment because there are two possible outcomes when customers enter the store — they either make a purchase or they do not.

Solve the following questions to help Izabella and Dylan find out if they can use their knowledge about binomial distributions in the store.

a Consider the following probability distributions.

Based on past experience, Magdalena knows that the probability that a customer will make a purchase is about

0.35 .

Suppose that

10

customers will enter the store in the next hour. Which graph describes the probability associated with the number of customers who will make a purchase?

b Suppose the probability that one customer will make a purchase changes to

0.5 .

What is the probability that exactly

3

of the next

8

customers will make a purchase? Round to two decimal places.

c Suppose that the store forecasts that

1500

customers will enter the store next month. If the probability that a customer will make a purchase is

0.35,

find the expected number of customers that make a purchase next month.

Hint

a Begin by identifying the conditions of a binomial experiment.

b Evaluate the Binomial Probability Formula at

x = 3 .

c The expected value is given by the product of the number of trials and the probability of success.

Solution

a Notice that the graphs each consider ten customers, so there are ten trials. In each trial, a success is that a customer makes a purchase. Start by identifying whether this experiment follows the three conditions of a binomial experiment.

Condition	Given Experiment	Is Satisfied?
There is a fixed number of independent trials.	There are $10$ trials — the $10$ customers that will enter the store.	$✓$
Each trial has two possible outcomes.	Whether a customer makes a purchase or not.	$✓$
The probability of success is constant for each trial.	There is a $0.35$ probability that a customer will make a purchase.	$✓$

Let

X

be the random variable representing the number of customers that will make a purchase. The possible values of

X

are

x = 0,

1,

\dots,

10 .

Note that the probability of failure

q

will be

1 - 0.35 = 0.65 .

This information can now be input into the Binomial Probability Formula.

P (X = x) =_{n} C_{x} p^{x} q^{n - x} ⇓ P (X = x) =_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}

The binomial probability distribution will be determined by evaluating this formula for each of the possible values of

x .

In this case, the probability that none of the customers will make a purchase will be found first.

P (X = x) =_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}

Substitute

$x = 0$

P (X = 0) =_{10} C_{0} (0.35)^{0} (0.65)^{10 - 0}

▼

Evaluate right-hand side

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

P (X = 0) = (\frac{1 0 !}{0 ! ( 1 0 - 0 ) !}) (0.35)^{0} (0.65)^{10 - 0}

SubTerms

Subtract terms

P (X = 0) = (\frac{1 0 !}{0 ! ( 1 0 ! )}) (0.35)^{0} (0.65)^{10}

$0! = 1$

P (X = 0) = (\frac{1 0 !}{1 ( 1 0 ! )}) (0.35)^{0} (0.65)^{10}

MultByOne

$a \cdot 1 = a$

P (X = 0) = (\frac{1 0 !}{1 0 !}) (0.35)^{0} (0.65)^{10}

QuotOne

$\frac{a}{a} = 1$

P (X = 0) = 1 (0.35)^{0} (0.65)^{10}

CalcPowProd

Calculate power and product

P (X = 0) = 0.013462 \dots

RoundDec

Round to $4$ decimal place(s)

P (X = 0) \approx 0.0135

Similarly, the probability for the remaining values of

x

will be found.

$x$	$_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}$	$P (X = x) =_{10} C_{x} (0.35)^{x} (0.65)^{10 - x}$
$0$	$_{10} C_{0} (0.35)^{0} (0.65)^{10 - 0}$	$P (X = 0) \approx 0.01346$
$1$	$_{10} C_{1} (0.35)^{1} (0.65)^{10 - 1}$	$P (X = 1) \approx 0.07249$
$2$	$_{10} C_{2} (0.35)^{2} (0.65)^{10 - 2}$	$P (X = 2) \approx 0.17565$
$3$	$_{10} C_{3} (0.35)^{3} (0.65)^{10 - 3}$	$P (X = 3) \approx 0.25222$
$4$	$_{10} C_{4} (0.35)^{4} (0.65)^{10 - 4}$	$P (X = 4) \approx 0.23767$
$5$	$_{10} C_{5} (0.35)^{5} (0.65)^{10 - 5}$	$P (X = 5) \approx 0.15357$
$6$	$_{10} C_{6} (0.35)^{6} (0.65)^{10 - 6}$	$P (X = 6) \approx 0.06891$
$7$	$_{10} C_{7} (0.35)^{7} (0.65)^{10 - 7}$	$P (X = 7) \approx 0.02120$
$8$	$_{10} C_{8} (0.35)^{8} (0.65)^{10 - 8}$	$P (X = 8) \approx 0.00428$
$9$	$_{10} C_{9} (0.35)^{9} (0.65)^{10 - 9}$	$P (X = 9) \approx 0.00051$
$10$	$_{10} C_{10} (0.35)^{10} (0.65)^{10 - 10}$	$P (X = 10) \approx 0.00003$

The table can now be used to graph the probability distribution.

The graphed distribution corresponds to option C.

b In order to find the probability that exactly

3

out of

8

customers will make a purchase,

P (X = 3)

needs to be found. In this case, the number of trials is

8,

the probability of success is

0.5,

and the probability of failure is

0.5 .

Use the Binomial Probability Formula again.

P (X = x) =_{8} C_{x} (0.5)^{x} (0.5)^{8 - x}

Substitute

$x = 3$

P (X = 3) =_{8} C_{3} (0.5)^{3} (0.5)^{8 - 3}

▼

Evaluate right-hand side

$C (n, k) = \frac{n !}{k ! ( n - k ) !}$

P (X = 3) = (\frac{8 !}{3 ! \cdot ( 8 - 3 ) !}) (0.5)^{3} (0.5)^{8 - 3}

SubTerms

Subtract terms

P (X = 3) = (\frac{8 !}{3 ! \cdot 5 !}) (0.5)^{3} (0.5)^{5}

MultPow

$a^{m} \cdot a^{n} = a^{m + n}$

P (X = 3) = (\frac{8 !}{3 ! \cdot 5 !}) (0.5)^{8}

SplitIntoFactors

Split into factors

P (X = 3) = (\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2 \cdot 1 \cdot 5 !}) (0.5)^{8}

CancelCommonFac

Cancel out common factors

P (X = 3) = (\frac{8 \cdot 7 \cdot 6 \cdot 5 !}{3 \cdot 2 \cdot 1 \cdot 5 !}) (0.5)^{8}

SimpQuot

Simplify quotient

P (X = 3) = (\frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1}) (0.5)^{8}

Multiply

P (X = 3) = (\frac{3 3 6}{6}) (0.5)^{8}

CalcQuot

Calculate quotient

P (X = 3) = 56 (0.5)^{8}

CalcPowProd

Calculate power and product

P (X = 3) = 0.21875

RoundDec

Round to $2$ decimal place(s)

P (X = 3) \approx 0.22

There is a probability of about

0.22

that exactly

3

out of

8

customers will make a purchase.

c Magdalena estimates that about

1500

customers will enter the store next month. Let

X

be the random variable representing the number of customers that will make a purchase. Since this random variable follows a binomial distribution, its expected value is given by the product of the number of trials

n

and the probability of success

p .

\underline{Expected Value of X} E (X) = n p

In this case, the number of trials

n

1500

and the probability of success is

0.35 .

The expected number of customers who will make a purchase next month can be determined with this information.

E (X) = 1500 \cdot 0.35 ⇕ E (X) = 525

Therefore,

525

customers are expected to make a purchase next month.

Closure

Normal Approximation to the Binomial Distribution

The binomial distribution is a discrete distribution that helps find the probability of the number of successes in a binomial experiment. In some cases, these calculations may be too complex. In such instances, a continuous distribution called normal distribution might approximate the calculations.

This graph shows a normal distribution curve above a number line. The curve is symmetrical with a peak at 950, indicating the mean value. The distribution is plotted above a number line, indicating that the values range approximately from 875 to 1025 in increments of 25.

The applet below shows both distributions when the probability of success is

0.5 .

approximation of binomial distribution using normal

The normal distribution graph is bell-shaped and symmetrical around the mean, its highest point. Additionally, if the number of trials increases and the probability of success is closer to

0.5,

the binomial distribution will closely resemble the normal distribution. The main characteristics of the normal distribution will be presented in another lesson.

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Catch-Up and Review

Example

Answer

Hint

Solution

Example

Variance vs Standard Deviation

Hint

Solution

Collection I

Collection II

Conclusion

Example

Extra

Hint

Solution

The Scratch-off Cards

Heights of 100 People

Rolling a Die

Blood Type O

Conclusion

Example

Extra

Proof

The Probability for a Fixed Sequence

Finding the Total Number of Possible Sequences

Getting the Binomial Probability

Hint

Solution

Hint

Solution

Heights of $100$ People