8C Preview

Preparing for the next class
In the next in-class activity, you will need to identify whether a chance experiment is a binomial experiment, determine the probability distribution for binomial experiments where [latex]n[/latex] is small, and understand the basic principles behind the formula for the binomial probability model.
A Bernoulli trial is a chance experiment with the following three properties

  • There are exactly two possible outcomes of the chance experiment. We label one of them as a success and the other as a failure (these aren’t value judgments on the outcomes, just labels; usually, we call the outcome we’re most interested in the success outcome).
  • The probability of success is the same for every trial. We call the probability of success [latex]p[/latex]. Since the only two outcomes are success and failure, the probability of failure is the probability that the trial does not result in a success, so we can use the NOT probability rule from In-Class Activity 7.B to find that the probability of failure is [latex]1-p[/latex].
  • The trials are independent from one another. (Recall from In-Class Activity 7.C that this means that the outcome of one trial does not affect the likelihood of the possible outcomes of subsequent trials.)

A binomial experiment is an experiment consisting of a fixed number, [latex]n[/latex], of independent Bernoulli trials that counts the number of successes out of [latex]n[/latex]trials. Notice that the number of successes in a binomial experiment is a discrete random variable. The distribution of this random variable is modeled with the binomial distribution.

Question 1

Determine whether each of the following is a binomial experiment, i.e., can we model the situation with the binomial distribution?

  1. A standard deck of 52 cards contains four aces. Three cards are drawn from the deck without replacement, and the number of aces that are drawn is recorded.
    1. This is a binomial experiment.
    2. This is not a binomial experiment because the number of trials is not fixed.
    3. This is not a binomial experiment because the probability of success is not the same for each trial.
    4.  This is not a binomial experiment because each trial has more than two outcomes.
  2. The number of people with blood type O-negative from a simple random sample of American adults of size 10 is recorded. According to the American Red Cross[1], seven percent of people in the United States have blood type O-negative.
    1. This is a binomial experiment.
    2. This is not a binomial experiment because the number of trials is not fixed.
    3. This is not a binomial experiment because the probability of success is not the same for each trial.
    4.  This is not a binomial experiment because each trial has more than two outcomes.
      Note: Even if the sample is drawn without replacement, the population of American adults is so large compared to the sample size of 10 that the probability of success is very close to the same for each trial, so we can consider the trials to be independent.
  3. Roll a six-sided die and count how many rolls it takes to roll a 4.
    1. This is a binomial experiment.
    2. This is not a binomial experiment because the number of trials is not fixed.
    3. This is not a binomial experiment because the probability of success is not the same for each trial.
    4.  This is not a binomial experiment because each trial has more than two outcomes.
  4. A simple random sample of American registered voters is taken, and the numbers of people who identify as Democrats, Republicans, and Third Party are recorded.
    1. This is a binomial experiment.
    2. This is not a binomial experiment because the number of trials is not fixed.
    3. This is not a binomial experiment because the probability of success is not the same for each trial.
    4.  This is not a binomial experiment because each trial has more than two outcomes.
  5. Roll a pair of six-sided dice 20 times and count how many times doubles were rolled, where a “double” means that the same number is rolled on both dice.
    1. This is a binomial experiment.
    2. This is not a binomial experiment because the number of trials is not fixed.
    3. This is not a binomial experiment because the probability of success is not the same for each trial.
    4.  This is not a binomial experiment because each trial has more than two outcomes.

Flipping a coin is a classic example of a Bernoulli trial. When we flip a coin, there are two possible outcomes, heads or tails, and the different flips of the coin are independent. We can think of flipping tails as a “success,” and if the coin is fair, the probability of success is [latex]p[/latex] = 0.5 for every trial.

Recall from Preview Assignment 8.B the experiment of flipping a coin 3 times and counting the number of tails obtained. We will revisit this example in the context of a binomial distribution.

Flipping a coin 3 times is a binomial experiment where [latex]n[/latex]= 3 and [latex]p[/latex]= 0.5, and the random variable [latex]X[/latex] is the number of tails in 3 coin flips. Therefore, the distribution of [latex]X[/latex] can be modeled using the binomial distribution. Recall that the outcomes of the experiment are as given in the following table:

Experimental Outcome [latex]X[/latex], Number of Tails in 3 Flips of a Coin
HHH 0
HHT 1
HTH 1
THH 1
TTH 2
THT 2
HTT 2
TTT 3

In Preview Assignment 8.B, you found the probability of each value of [latex]X[/latex] by looking at relative frequencies. Notice that since the trials here are independent, you can also find these probabilities by using the rule for independent events, [latex]P(A \mbox{ and } B) = P(A) P(B)[/latex], in combination with the rule for finding OR probabilities.

Question 2

Find the following probabilities using the rule for independent events.

  1. Find [latex]P[/latex](HHT) =
  2. Find [latex]P[/latex] (HTH) =
  3. Use the previous chart to help you complete the following table describing the number of ways to obtain x tails in 3 coin flips.
  4. Number of Tails, [latex]x[/latex] Ways to Obtain [latex]x[/latex] Tails Number of Ways to Obtain [latex]x[/latex] Tails in 3 Flips
    0 HHH 1
    1 THH, HTH, HHT 3
    2
    3

Notice that the probabilities you found in Parts a and b are the same. This is due to the fact that the trials are independent!

Recall from In-Class Activity 7.B that [latex]P(A~or~B) =P(A)+P(B)[/latex] if the events are mutually exclusive (meaning they cannot happen at the same time).

Since the outcomes THH, HTH, and HHT are mutually exclusive, the probability of obtaining 1 tail in 3 coin flips, i.e., [latex]P(X = 1)[/latex] , is:

[latex]P(X = 1) = P(THH) + P(HTH) + P(HHT) = 0.125 + 0.125 + 0.125 = 0.375[/latex]

Since all 3 of the outcomes yielding 1 tail have the same probability, then

[latex]P(THH) = P(HTH) = P(HHT) = P(T)P(H)P(H)=P(T)(P(H))^{2}[/latex]

and we can compute the probability [latex]P(X = 1)[/latex] as follows:

 

[latex]P(X=1) = 3 P(T)P(H)P(H) = 3P(T)(P(H))^{2} = 3(0.5)(0.5)^{2} = 0.375[/latex]

 

where, as you observed in the table, 3 is the number of ways of obtaining 1 tail in 3 coin flips.

 

This example gives us a glimpse into the formula for the binomial distribution, which is used to model a binomial experiment.

In general, the formula for the probability of obtaining x successes from n independent trials where the probability of success is p is:

 

[latex]P(X = x) = (number~of~ways~to~obtain~x~successes~in~n~trials) \bullet p^{x} \bullet (1-p)^{n-x}[/latex]

where [latex]p^{x}[/latex] occurs because there are x successes, and [latex](1-p)^{n-x}[/latex] occurs because if there are [latex]x[/latex] successes and [latex]n[/latex] trials total, there must be [latex]n-x[/latex] failures.

Question 3

Now, let’s make things a bit more interesting and imagine that our coin is not fair. In fact, it lands on tails 70% of the time. Consider the binomial experiment of flipping the coin 3 times and counting how many times tails lands face up, so X is the number of tails obtained in 3 coin flips.

  1. What are the probability of success ([latex]p[/latex]) and the probability of failure [latex](1-p)[/latex] for our new experiment?
  2. Complete the following table describing the probability distribution of X. Use what you learned in Question 2, where a similar binomial experiment was explored but the coin was fair.For example, to find the probability of obtaining 1 tail in 3 flips, we would compute:[latex]P(X=1) = P(THH~or~HTH~or~HHT) =3P(T)(P(H))^{2} = 3p(1-p)^{2} = 0.189[/latex]
    x [latex]P(X=x)[/latex]
    0 [latex]1(0.7)^{0}(0.3)^{3} = 0.027[/latex]
    1 0.189
    2
    3

Looking Ahead

Question 4

Notice that the number of ways to obtain each number of successes in a binomial experiment increases pretty quickly. If we were to flip 4 coins, there would be:

  • 1 way to obtain 0 tails
  • 4 ways to obtain 1 tail
  • 6 ways to obtain 2 tails
  • 4 ways to obtain 3 tails
  • 1 way to obtain 4 tails

There is a formula that lets us compute these probabilities more easily. For a binomial experiment in which the probability of success is p and there are n

trials, the binomial distribution gives the probability of obtaining x successes as

 

[latex]P(X=x) = \frac{n!}{x!(n-x)!} \bullet p^{x} \bullet (1-p)^{n-x}[/latex]

 

where [latex]\frac{n!}{x!(n-x)!}[/latex] is called “[latex]n \mbox{ choose } x[/latex],” which computes the number of ways to obtain [latex]x[/latex] successes out of [latex]n[/latex] trials.

The exclamation mark is the symbol for a factorial. You won’t need to calculate this because we will be using technology for our computations, but [latex]n![/latex] is the product

 

[latex]n! = n(n-1)(n-2) \cdots (2)(1)[/latex]

 

For example, [latex]3! =(3)(2)(1) = 6[/latex].

We found that there were 3 ways to obtain 1 tail in 3 coin flips. To see how this corresponds to the formula, observe that

 

[latex]3~choose~1 = \frac{3!}{1(!3-1!)} = \frac{3!}{1!2!} = \frac{6}{1 \bullet 2} = \frac{6}{2} = 3[/latex]

 

As mentioned, we will be using technology to compute these probabilities, so you won’t need to worry much about the formula. We will be using the Binomial Distribution tool found at https://dcmathpathways.shinyapps.io/BinomialDist/.

Feel free to check out the tool before class! You can click on the Find Probabilities tab, input values for [latex]n, p, \mbox{ and } x[/latex], and then select which type of probability you would like to compute from the drop-down menu.

 

  1. American Red Cross. (n.d.). Facts about blood and blood types. https://www.redcrossblood.org/donate blood/blood-types.html ↵