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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**

**Understanding Descriptive Measures**

**Understanding Types of Data**

**Other Recommended Readings**

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.

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?

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*.

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 $1.$

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.

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, because rolling each die has $6$ possible outcomes, the number of all possible results is $6⋅6=36.$ 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 | $P(X=x)$ |

$2$ | $1$ | $361 ≈0.028$ |

$3$ | $2$ | $362 ≈0.056$ |

$4$ | $3$ | $363 ≈0.083$ |

$5$ | $4$ | $364 ≈0.111$ |

$6$ | $5$ | $365 ≈0.139$ |

$7$ | $6$ | $366 ≈0.167$ |

$8$ | $5$ | $365 ≈0.139$ |

$9$ | $4$ | $364 ≈0.111$ |

$10$ | $3$ | $363 ≈0.083$ |

$11$ | $2$ | $362 ≈0.056$ |

$12$ | $1$ | $361 ≈0.028$ |

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

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 $X.$

a

$X=Number of Heads$ | |||||
---|---|---|---|---|---|

$x$ | $0$ | $1$ | $2$ | $3$ | $4$ |

$P(X=x)$ | $0.0625$ | $0.25$ | $0.375$ | $0.25$ | $0.0625$ |

b

c

$X=Number of Heads$ | |||||
---|---|---|---|---|---|

$x$ | $0$ | $1$ | $2$ | $3$ | $4$ |

$P(X=x)$ | $0.06$ | $0.22$ | $0.37$ | $0.28$ | $0.07$ |

d

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

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.

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 | $TTTT$ | $TTTH,$ $TTHT,$ $THTT,$ $HTTT$ | $HHTT,$ $HTTH,$ $TTHH,$ $HTHT,$ $THTH,$ $THHT$ | $HHHT,$ $HHTH,$ $HTHH,$ $THHH$ | $HHHH$ |

Frequency | $1$ | $4$ | $6$ | $4$ | $1$ |

$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 | $1$ | $4$ | $6$ | $4$ | $1$ |

$P(X=x)$ | $161 $ | $164 =41 $ | $166 =83 $ | $164 =41 $ | $161 $ |

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 | $P(X=x)$ |
---|---|---|---|

$0$ | $∣∣∣∣ ∣$ | $6$ | $1006 $ |

$1$ | $∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣$ | $22$ | $10022 $ |

$2$ | $∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣$ | $37$ | $10037 $ |

$3$ | $∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣∣ ∣∣∣$ | $28$ | $10028 $ |

$4$ | $∣∣∣∣ ∣∣$ | $7$ | $1007 $ |

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.

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.

$E(X)=i=1∑n x_{i}⋅P(X=x_{i})$

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.

Let $X_{1},$ $X_{2},$ $…,$ $X_{n}$ be independent random variables with an identical probability distribution and $S_{n} $ be the mean of the random variables.
This notation means that for a sufficiently large number of observations $n$ — or by repeating the experiment a large number of times — the mean $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.

$S_{n} =nX_{1}+X_{2}+…+X_{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 $S_{n} $ tends towards $μ.$ $n→+∞lim S_{n} =μ$

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 $0.5.$

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.

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.

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.

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 $E(X).$
The variance is denoted by $σ_{2}$ because it is the square of the standard deviation $σ.$

$Deviation ofx_{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}⋅P(X=x_{i})] $

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}⋅P(X=x_{i})] $

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

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 $X.$

$x$ | $0$ | $1$ | $2$ | $3$ | $4$ |
---|---|---|---|---|---|

$P(X=x)$ | $151 $ | $153 $ | $156 $ | $153 $ | $152 $ |

$E(X)=[x_{1}⋅P(X=x_{1})]+[x_{2}⋅P(X=x_{2})]+…+[x_{n}⋅P(X=x_{n})]$

SubstituteValues

Substitute values

$E(X)=0(151 )+1(153 )+2(156 )+3(153 )+4(152 )$

Evaluate right-hand side

MoveLeftFacToNum

$a⋅cb =ca⋅b $

$E(X)=150 +153 +1512 +159 +158 $

$a0 =0$

$E(X)=0+153 +1512 +159 +158 $

IdPropAdd

Identity Property of Addition

$E(X)=153 +1512 +159 +158 $

AddFrac

Add fractions

$E(X)=1532 $

UseCalc

Use a calculator

$E(X)=2.1333…$

RoundDec

Round to $2$ decimal place(s)

$E(X)≈2.13$

$x_{i}$ | $[x_{i}−E(X)]_{2}$ | $[x_{i}−E(X)]_{2}⋅P(X=x_{i})$ |
---|---|---|

$0$ | $(0−2.13)_{2}=4.5369$ | $4.5364⋅151 ≈0.3025$ |

$1$ | $(1−2.13)_{2}=1.2769$ | $1.2769⋅153 ≈0.2554$ |

$2$ | $(2−2.13)_{2}=0.0169$ | $0.0169⋅156 ≈0.0068$ |

$3$ | $(3−2.13)_{2}=0.7569$ | $0.7569⋅153 ≈0.1514$ |

$4$ | $(4−2.13)_{2}=3.4969$ | $3.4969⋅152 ≈0.4663$ |

Variance $σ_{2}$ | $≈1.1824$ |

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

$Standard Deviation: 1.1824 ≈1.09 $

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.

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

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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.

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.

$E(X)=[x_{1}⋅P(X=x_{1})]+[x_{2}⋅P(X=x_{2})]+…+[x_{n}⋅P(X=x_{n})]$

SubstituteValues

Substitute values

$E(X)=1200⋅0.5+1800⋅0.2+900⋅0.2+(-150)⋅0.1$

Multiply

Multiply

$E(X)=600+360+180−15$

AddSubTerms

Add and subtract terms

$E(X)=1125$

$σ=Σ[[x_{i}−E(X)]_{2}⋅P(X=x_{i})] $

To apply this formula, calculate the square of each $x_{i}$ | $[x_{i}−E(X)]_{2}$ | $[x_{i}−E(X)]_{2}⋅P(X=x_{i})$ |
---|---|---|

$1200$ | $(1200−1125)_{2}=5625$ | $5625⋅0.5=2812.5$ |

$1800$ | $(1800−1125)_{2}=455625$ | $455625⋅0.2=91125$ |

$900$ | $(900−1125)_{2}=50625$ | $50625⋅0.2=10125$ |

$-150$ | $(-150−1125)_{2}=1625625$ | $1625625⋅0.1=162562.5$ |

Sum of Values | $266625$ |

$Standard Deviation: 266625 ≈516.36 $

$E(X)=[x_{1}⋅P(X=x_{1})]+[x_{2}⋅P(X=x_{2})]+…+[x_{n}⋅P(X=x_{n})]$

SubstituteValues

Substitute values

$E(X)=3600⋅0.3+2850⋅0.1+(-300)⋅0.4+(-500)⋅0.2$

Multiply

Multiply

$E(X)=1080+285−120−100$

AddSubTerms

Add and subtract terms

$E(X)=1145$

$x_{i}$ | $[x_{i}−E(X)]_{2}$ | $[x_{i}−E(X)]_{2}⋅P(X=x_{i})$ |
---|---|---|

$3600$ | $(3600−1145)_{2}=6027025$ | $6027025⋅0.3=1808107.5$ |

$2850$ | $(2850−1145)_{2}=2907025$ | $2907025⋅0.1=290702.5$ |

$-300$ | $(-300−1145)_{2}=2088025$ | $2088025⋅0.4=835210$ |

$-500$ | $(-500−1145)_{2}=2706025$ | $2706025⋅0.2=541205$ |

Sum of Values | $3475225$ |

$Standard Deviation: 3475225 ≈1864.20 $

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 | $≈516.23$ | |

Standard Deviation of Collection II | $≈1864.20$ |

b Consider the expected value of each distribution.

$Expected Values Collection I:Collection II: E(X)=1125E(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. $Standard Deviation Collection I:Collection II: σ≈516.36σ≈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.
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*.

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.

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

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.

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

### Solution

### The Scratch-off Cards

### Heights of $100$ People

### Rolling a Die

In this case, rolling a $3$ can take only one or many more trials. ### Blood Type O

### Conclusion

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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?

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.

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 $25%$ or $0.25,$ 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. 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.

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.

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. 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 | $×$ |

Rolling a Die | $×$ |

Blood Type O? | $✓$ |

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.

The binomial distribution is the probability distribution that describes the number of successes $x$ out of $n$ binomial trials. The trials must follow these 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 $x=0,$ $1,$ $2,$ $…,$ $n.$ 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 $p.$

$E(X)=np$

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

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 $5213 =41 .$ The probability of failure $q$ will be $1−41 =43 .$ For example, consider repeating this experiment $5$ times. Then, the binomial distribution with $n=5$ can be determined.$P(X=x)=_{5}C_{x}(41 )_{x}(43 )_{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. $x$ | $_{5}C_{x}(41 )_{x}(43 )_{5−x}$ | $P(X=x)=_{5}C_{x}(41 )_{x}(43 )_{5−x}$ |
---|---|---|

$0$ | $_{5}C_{0}(41 )_{0}(43 )_{5−0}=1024243 $ | $≈0.237$ |

$1$ | $_{5}C_{1}(41 )_{1}(43 )_{5−1}=1024405 $ | $≈0.396$ |

$2$ | $_{5}C_{2}(41 )_{2}(43 )_{5−2}=1024270 $ | $≈0.264$ |

$3$ |