7.3: Binomial Distribution and Binomial Experiments
Learning Objectives
 Apply techniques to estimate the probability of a population proportion of outcomes in a survey and experiment.
 Understand the conditions needed to conduct a binomial experiment.
Introduction
A probability distribution shows the possible numerical outcomes of a chance process and from this the probability of any set of possible outcomes can be deduced. As seen in previous lessons, for many events probability distribution can be modeled by the normal curve. One type of probability distribution that is worth examining is a binomial distribution.
In probability theory and statistics, the binomial distribution is the discrete probability distribution of the number of successes in a sequence of “\begin{align*}n\end{align*}
To conduct a binomial experiment a random sample (\begin{align*}n\end{align*}
 The number of people with type \begin{align*}\mathrm{O}\end{align*}
O blood in a random sample of \begin{align*}10\end{align*}10 people (a person either has type \begin{align*}\mathrm{O}\end{align*}O blood or doesn’t.)  The number of doubles in eight rolls of a pair of dice (doubles either show up on each roll or they don’t.)
 The number of defective light bulbs in a sample of \begin{align*}30\end{align*}
30 bulbs (either the bulb is defective or it isn’t.)
Binomial Experiments
These events are called binomial because each one has two possible outcomes. Let’s examine an actual binomial situation. Suppose we present four people with two cups of coffee (one percolated and one instant) to discover the answer to this question: “If we ask four people which is percolated coffee and none of them can tell the percolated coffee from the instant coffee, what is the probability that two of the four will guess correctly?” We will present each of four people with percolated and instant coffee and ask them to identify the percolated coffee. The outcomes will be recorded by using \begin{align*}C\end{align*}
Number Who Correctly Identify Percolated Coffee 
Outcomes
\begin{align*}C\end{align*}
\begin{align*}I\end{align*} 
Number of Outcomes 

\begin{align*}IIII\end{align*} 
\begin{align*}1\end{align*} 

\begin{align*}1\end{align*} 
\begin{align*}ICII\ IIIC\ IICI\ CIII\end{align*} 
\begin{align*}4\end{align*} 
\begin{align*}2\end{align*} 
\begin{align*}ICCI\ IICC\ ICIC\ CIIC\end{align*} 
\begin{align*}6\end{align*} 
\begin{align*}3\end{align*} 
\begin{align*}CICC\ ICCC\ CCCI\ CCIC\end{align*} 
\begin{align*}4\end{align*} 
\begin{align*}4\end{align*} 
\begin{align*}CCCC\end{align*} 
1 
Using the Multiplication Rule for Independent Events, you know that the probability of getting a certain outcome when two people guess correctly, like, \begin{align*}CICI\end{align*}
\begin{align*}_4 C _2 = \frac{4!} {2! 2!} = \frac{24} {4} = 6\end{align*}
A graphing calculator can also be used to calculate binomial probabilities.
\begin{align*}2^{nd}\end{align*}
A random sample can be treated as a binomial situation if the sample size, \begin{align*}n\end{align*}
A binomial experiment is a probability experiment that satisfies the following conditions:
 Each trial can have only two outcomes – one known as “success” and the other “failure.”
 There must be a fixed number, \begin{align*}n\end{align*}
n , of trials.  The outcomes of each trial must be independent of each other. The probability of each a “success” doesn’t change regardless of what occurred previously.
 The probability, \begin{align*}p\end{align*}
p , of a success must remain the same for each trial.
The distribution of the random variable \begin{align*}X\end{align*}
\begin{align*}P(X = k) = \binom{n}{k} p^k (1  p)^{n  k}\end{align*}
where \begin{align*}\binom{n}{k} = \frac{n!} {k!(n  k)!}\end{align*}
Let’s return to the coffee experiment and look at the distribution of \begin{align*}X\end{align*} (correct guesses):
\begin{align*}& k && P(X = k)\\ & 0 && 1/16\\ & 1 && 4/16\\ & 2 && 6/16\\ & 3 && 4/16\\ & 4 && 1/16\end{align*}
The expected value for the above distribution is:
\begin{align*}E (X) = 0(1/16) + 1(4/16) + 2(6/16) + 3(4/16) + 4(1/16)\end{align*}
\begin{align*}E (X) = 2\end{align*} In other words, you expect half of the four to guess correctly when given two equally, likely choices. \begin{align*}E(X)\end{align*} can be written as \begin{align*}4 \cdot \frac{1} {2}\end{align*} which is equivalent to \begin{align*}np.\end{align*} For a random variable \begin{align*}X\end{align*} having a binomial distribution with \begin{align*}n\end{align*} trials and probability of successes \begin{align*}p\end{align*}, the expected value (mean) and standard deviation for the distribution can be determined by:
\begin{align*}E(X) = np = \mu_x && \text{and} && \sigma_x = \sqrt{np(1 p)}\end{align*}
The graphing calculator will now be used to graph and compare binomial distributions. The binomial distribution will be entered into two lists and then displayed as a histogram. First we will use the calculator to generate a sequence of integers and secondly the list of binomial probabilities.
Sequence of integers:
\begin{align*}2^{nd}\end{align*} [LIST] \begin{align*}\rightarrow\end{align*} OPS \begin{align*}\downarrow\end{align*} \begin{align*}5.\end{align*}seq ( ) STO \begin{align*}\rightarrow\end{align*} \begin{align*}2^{nd}\ 1\end{align*}
Binomial Probabilities:
\begin{align*}2^{nd}\end{align*} DISTR :binompdf(
Horizontally, the following are examples of binomial distributions where \begin{align*}n\end{align*} increases and \begin{align*}p\end{align*} remains constant. Vertically, the examples display the results where \begin{align*}n\end{align*} remains fixed and \begin{align*}p\end{align*} increases.
\begin{align*}n = 5 \ \ \text{and} \ \ p = 0.1 && n = 10 \ \ \text{and} \ \ p = 0.1 && n = 20 \ \ \text{and} \ \ p = 0.1\end{align*}
For the small value of \begin{align*}p\end{align*}, the binomial distributions are skewed toward the higher values of \begin{align*}x\end{align*}. As \begin{align*}n\end{align*} increases, the skewness decreases and the distributions gradually move toward being more normal.
\begin{align*}n = 5 \ \ \text{and} \ \ p = 0.5 && n = 10 \ \ \text{and} \ \ p = 0.5 && n = 20 \ \ \text{and} \ \ p = 0.5\end{align*}
As \begin{align*}p\end{align*} increases to \begin{align*}0.5\end{align*}, the skewness disappeared and the distributions achieved perfect symmetry. The symmetrical, moundshaped distribution remained the same for all values of \begin{align*}n\end{align*}.
\begin{align*}n = 5 \ \ \text{and} \ \ p = 0.75 && n = 10 \ \ \text{and} \ \ p = 0.75 && n = 20 \ \ \text{and} \ \ p = 0.75\end{align*}
For the larger value of \begin{align*}p\end{align*}, the binomial distributions are skewed toward the lower values of \begin{align*}x\end{align*}. As \begin{align*}n\end{align*} increases, the skewness decreases and the distributions gradually move toward being more normal.
Because \begin{align*}E(X) = np = \mu_x\end{align*}, the value increases with both \begin{align*}n\end{align*} and \begin{align*}p\end{align*}. As \begin{align*}n\end{align*} increases, so does the standard deviation but for a fixed value of \begin{align*}n\end{align*}, the standard deviation is largest around \begin{align*}p = 0.5\end{align*} and reduces as \begin{align*}p\end{align*} approaches or \begin{align*}1\end{align*}.
Example: Suppose you flip a fair coin \begin{align*}12 \;\mathrm{times}\end{align*}.
a) What is the probability that you will get exactly \begin{align*}5\end{align*} heads?
b) Exactly \begin{align*}25\%\end{align*} heads?
c) At least \begin{align*}10\end{align*} heads?
Solution: Let \begin{align*}X\end{align*} represent the number of heads from \begin{align*}12\end{align*} flips of the coin. Exactly \begin{align*}5\end{align*} heads:
a)
\begin{align*}\binom{n}{k} & = \frac{n!} {k!(n  k)!} & & P(X = 5) = 792(0.5)^{12} = 0.1933\\ \binom{12}{5} & = \frac{12!} {5!(12  5)!}\\ \binom{12}{5} & = 792\end{align*}
b) \begin{align*}25\%\end{align*} of \begin{align*}12\end{align*} heads means getting \begin{align*}3\end{align*} heads.
\begin{align*}\binom{n}{k} & = \frac{n!} {k!(n  k)!} & & P(X = 3) = 220(0.5)^{12} = 0.0537\\ \binom{12}{3} & = \frac{12!} {3!(12  3)!}\\ \binom{12}{3} & = 220\end{align*}
c) At least \begin{align*}10\end{align*} heads means getting \begin{align*}10\end{align*} heads, \begin{align*}11\end{align*} heads or \begin{align*}12\end{align*} heads.
\begin{align*}& \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{12}{10} = \frac{12!} {10!(12  10)!} & & \binom{12}{10} = 66\\ & \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{12}{11} = \frac{12!} {11!(12  11)!} & & \binom{12}{11} = 12\\ & \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{12}{12} = \frac{12!} {12!(12  12)!} & & \binom{12}{12} = 1\end{align*}
\begin{align*}& P(X = 10) = 66(0.5)^{12} \approx 0.0161\\ & P(X = 11) = 12(0.5)^{12} \approx 0.0029\\ & P(X = 12) = 1(0.5)^{12} \approx 0.0002\end{align*}
At least \begin{align*}10\end{align*} heads: \begin{align*}0.0161 + 0.0029 + 0.0002 \approx 0.0192\end{align*}
Example:
Since the closing of the two main industries, a small town has experienced a decrease in the population of teenagers. According to the statistics of the local high school, approximately \begin{align*}7.6\%\end{align*} of teens ages \begin{align*}14  19\end{align*} are no longer registered. Suppose you randomly chose four people who were enrolled at the high school before the population dropped.
a) What is the probability that none of the four are registered?
b) What is the probability that at least one is not registered?
c) Create a probability distribution table for this situation.
Solution:
a)
\begin{align*}& P (X = k) = \binom{n}{k} P^k (1  p)^{n  k}\\ & P(X = 0) \binom{4}{0} (0.076)^0 (1  0.076)^4 = 0.7289\end{align*}
b)
\begin{align*}& P(X \ge 1) = 1  [P(X = 0)]\\ & P(X \ge 1) \approx 1  0.7289 = 0.2711\end{align*}
c) Using technology:
If you so desire, you can transfer this data into your own table:
Number Not Registered  Probability 

\begin{align*}0.72893\end{align*}  
\begin{align*}1\end{align*}  \begin{align*}0.23982\end{align*} 
\begin{align*}2\end{align*}  \begin{align*}0.02959\end{align*} 
\begin{align*}3\end{align*}  \begin{align*}0.00162\end{align*} 
\begin{align*}4\end{align*}  \begin{align*}0.00003\end{align*} 
Lesson Summary
In this lesson you have learned that the random variable \begin{align*}X\end{align*} has a binomial distribution if \begin{align*}X\end{align*} represents the number of “successes” in \begin{align*}n\end{align*} independent trials. In each of these trials, the probability of success is \begin{align*}p\end{align*}. You have also learned that a binomial distribution has these important features:
 The probability of getting exactly \begin{align*}X = k\end{align*} successes is given by the formula
\begin{align*}P(X = k) = \binom{n}{k} p^k (1  p)^{n  k}\end{align*}
 The mean or expected value is represented by
\begin{align*}E(X) = np = \mu_x\end{align*}
 The standard deviation is
\begin{align*}\sigma_x = \sqrt{np (1  p)}\end{align*}
Review Questions
 According to the Canadian census of 2006, the median annual family income for families in Nova Scotia is \begin{align*}\$56,400\end{align*}. [Source: Stats Canada. www.statcan.ca ]. Consider a random sample of \begin{align*}24\end{align*} Nova Scotia households.
 What is the expected number of households with annual incomes less than \begin{align*}\$56,400\end{align*}?
 What is the standard deviation of households with incomes less than \begin{align*}\$56,400\end{align*}?
 What is the probability of getting at least \begin{align*}18\end{align*} out of the \begin{align*}24\end{align*} households with annual incomes under \begin{align*}\$56,400\end{align*}?
Review Answers
 (a) The operative word in this problem is ‘median’; if the median income is \begin{align*}\$56,400\end{align*}, then this indicates that onehalf of the income falls below \begin{align*}\$56,400\end{align*} and onehalf of the income lies above it. Therefore, the chance of a randomly selected income being below the median income is \begin{align*}0.5\end{align*}. Let \begin{align*}X\end{align*} represent the number of households with incomes below the median in a random sample of size \begin{align*}24\end{align*}. \begin{align*}X\end{align*} has a binomial distribution with \begin{align*}n = 24\end{align*} and \begin{align*}p = 0.5\end{align*}. \begin{align*}E(X) & = np = \mu_x\\ E(X)& = (24)(0.5) = 12\end{align*} (b) The standard deviation of households with incomes less than \begin{align*}\$56,400\end{align*} is \begin{align*}\sigma_x & = \sqrt{np (1  p)}\\ \sigma_x & = \sqrt{12 (1  0.5)}\\ \sigma_x & = 2.25\end{align*} (c) \begin{align*}P(X \ge 18)\approx\end{align*} \begin{align*}& \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{n}{k} = \frac{n!} {k!(n  k)!} \\ & \binom{24}{18} = \frac{24!} {18!(24  18)!} & & \binom{24}{19} = \frac{24!} {19!(24  19)!} & & \binom{24}{20} = \frac{24!} {20!(24  20)!}\\ & \binom{24}{18} = 134596 & & \binom{24}{19} = 42504 & & \binom{24}{20} = 10626\end{align*} \begin{align*}& \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{n}{k} = \frac{n!} {k!(n  k)!} & & \binom{n}{k} = \frac{n!} {k!(n  k)!}\\ & \binom{24}{21} = \frac{24!} {21!(24  21)!} & & \binom{24}{22} = \frac{24!} {22!(24  22)!} & & \binom{24}{23} = \frac{24!} {23!(24  23)!}\\ & \binom{24}{21} = 2024 & & \binom{24}{22} = 276 & & \binom{24}{23} = 24\end{align*} \begin{align*}& \binom{n}{k} = \frac{n!} {k!(n  k)!} && P(X = 18) = 134596(0.5)^{24} \approx 0.0080\\ & \binom{24}{24} = \frac{24!} {24!(24  24)!} & & P(X = 19) = 42504(0.5)^{24} \approx 0.0025\\ & \binom{24}{24} = 1 & & P(X = 20) = 10626(0.5)^{24} \approx 0.0006\\ & & & P(X = 21) = 2024(0.5)^{24} \approx 0.0001\\ & & & P(X = 22) = 276(0.5)^{24} \approx 0.000016\\ & & & P(X = 23) = 24(0.5)^{24} \approx 0.0000014\\ & & & P(X = 24) = 1(0.5)^{24} \approx 0.00000006\end{align*} The sum of these probabilities gives the answer to the question: \begin{align*}0.01121746.\end{align*}
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