AOS4 Topic 1: Binomial Distribution

Bernoulli Sequences and the Binomial Probability Distribution

An experiment often consists of repeated trials, each of which may be considered as having only two possible outcomes. For example, when a coin is tossed, the two possible outcomes are ‘head’ and ‘tail’. When a die is rolled, the two possible outcomes are determined by the random variable of interest for the experiment. If the event of interest is a ‘six’, then the two outcomes are ‘six’ and ‘not a six’.

A Bernoulli sequence is the name used to describe a sequence of repeated trials with the following properties:

  • Each trial results in one of two outcomes, which are usually designated as either a success, \( S \), or a failure, \( F \).
  • The probability of success on a single trial, \( p \), is constant for all trials (and thus the probability of failure on a single trial is \( 1 - p \)).
  • The trials are independent (so that the outcome of any trial is not affected by the outcome of any previous trial).

    • The Binomial Probability Distribution

    The number of successes in a Bernoulli sequence of \( n \) trials is called a binomial random variable and is said to have a binomial probability distribution.

    For example, consider rolling a fair six-sided die three times. Let the random variable \( X \) be the number of 3s observed.

    Let \( T \) represent a 3, and let \( N \) represent not a 3. Each roll meets the conditions of a Bernoulli trial. Thus \( X \) is a binomial random variable.

    Binomial Probability Distribution

    Now consider all the possible outcomes from the three rolls and their probabilities.

    Outcome Number of 3s Probability
    TTT X = 3 \( \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \)
    TTN X = 2 \( \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} \)
    TNT X = 2 \( \frac{1}{6} \times \frac{5}{6} \times \frac{1}{6} \)
    NTT X = 2 \( \frac{5}{6} \times \frac{1}{6} \times \frac{1}{6} \)
    TNN X = 1 \( \frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} \)
    NTN X = 1 \( \frac{5}{6} \times \frac{1}{6} \times \frac{5}{6} \)
    NNT X = 1 \( \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} \)
    NNN X = 0 \( \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \)

    Pr(X = 3) = \( \left(\frac{1}{6}\right)^3 \)

    Pr(X = 2) = \( 3 \times \left(\frac{1}{6}\right)^2 \times \frac{5}{6} \)

    Pr(X = 1) = \( 3 \times \frac{1}{6} \times \left(\frac{5}{6}\right)^2 \)

    Pr(X = 0) = \( \left(\frac{5}{6}\right)^3 \)

    Thus the probability distribution of \( X \) is given by the following table:

    x 0 1 2 3
    Pr(X = x) \( \frac{1}{216} \) \( \frac{15}{216} \) \( \frac{75}{216} \) \( \frac{125}{216} \)

    Consider the probability that X = 1, that is, when exactly one 3 is observed. We can see from the table that there are three ways this can occur. Since the 3 could occur on the first, second or third roll of the die, we can consider this as selecting one object from a group of three, which can be done in \( \frac{3}{1!} \) ways.

    Consider the probability that X = 2, that is, when exactly two 3s are observed. Again from the table there are three ways this can occur. Since the two 3s could occur on any two of the three rolls of the die, we can consider this as selecting two objects from a group of three, which can be done in \( \frac{3}{2!} \) ways.

    This leads us to a general formula for this probability distribution: \[ \text{Pr}(X = x) = \binom{3}{x} \left(\frac{1}{6}\right)^x \left(\frac{5}{6}\right)^{3-x} \quad \text{for} \quad x = 0, 1, 2, 3 \]

    This is an example of the binomial distribution.

    Binomial Distribution

    If the random variable X is the number of successes in \( n \) independent trials, each with probability of success \( p \), then \( X \) has a binomial distribution, written \( X \sim \text{Bi}(n, p) \), and the rule is:

    \[ \text{Pr}(X = x) = \binom{n}{x} p^x (1 - p)^{n-x} \quad \text{for} \quad x = 0, 1, \ldots, n \]

    where: \[ \binom{n}{x} = \frac{n!}{x!(n - x)!} \]

    The Binomial Distribution and Conditional Probability

    The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success, denoted by \( p \).

    Conditional probability, on the other hand, deals with the probability of an event occurring given that another event has already occurred. It is denoted by \( P(A | B) \), which represents the probability of event \( A \) occurring given that event \( B \) has already occurred.

    In the context of the binomial distribution, conditional probability can be used to calculate the probability of certain outcomes given specific conditions or information about previous outcomes.

    For example, if we have already observed a certain number of successes in a series of Bernoulli trials, conditional probability can be used to calculate the probability of observing additional successes or failures in subsequent trials.

    Graph of Binomial Probability Distribution

    A probability distribution may be represented as a rule, a table, or a graph. We now investigate the shape of the graph of a binomial probability distribution for different values of the parameters \( n \) and \( p \).

    Expectation and Variance of binomial distribution

    How many heads would you expect to obtain, on average, if a fair coin was tossed 10 times?

    While the exact number of heads in the 10 tosses would vary, and could theoretically take values from 0 to 10, it seems reasonable that the long-run average number of heads would be 5. It turns out that this is correct. That is, for a binomial random variable \( X \) with \( n = 10 \) and \( p = 0.5 \),

    \[ E(X) = \sum_{x=0}^{10} x \cdot \text{Pr}(X = x) = 5 \]

    In general, the expected value of a binomial random variable is equal to the number of trials multiplied by the probability of success. The variance can also be calculated from the parameters \( n \) and \( p \).

    Definition

    If \( X \) is the number of successes in \( n \) trials, each with a probability of success \( p \), then the expected value and the variance of \( X \) are given by:

    \[ E(X) = np \]

    \[ \text{Var}(X) = np(1 - p) \]

    Proofs for the expectation and variance

    The probabilities of a binomial distribution sum to 1.

    Proof: The binomial theorem, discussed in Appendix A, states that

    \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

    Now, using the binomial theorem, the sum of the probabilities for a binomial random variable \( X \) with parameters \( n \) and \( p \) is given by

    \[ \sum_{x=0}^{n} Pr(X = x) = \sum_{x=0}^{n} \binom{n}{x} p^x (1 - p)^{n-x} \]

    \[ = (1 - p) + p \sum_{x=0}^{n} \binom{n}{x} p^{x-1} (1 - p)^{(n-1)-(x-1)} \]

    \[ = (1 - p) + p \cdot 1^n = (1)^n = 1 \]

    Expected Value of a Binomial Random Variable

    If \( X \) is a binomial random variable with parameters \( n \) and \( p \), then \( E(X) = np \).

    Proof: By the definition of expected value:

    \[ E(X) = \sum_{x=0}^{n} x \cdot \binom{n}{x} p^x (1 - p)^{n-x} \] by the distribution formula

    \[ = \sum_{x=0}^{n} x \cdot \frac{n!}{x! (n - x)!} p^x (1 - p)^{n-x} \] expanding \(\binom{n}{x}\)

    \[ = \sum_{x=1}^{n} x \cdot \frac{n!}{x(x - 1)! (n - x)!} p^x (1 - p)^{n-x} \] since the \( x = 0 \) term is zero

    \[ = \sum_{x=1}^{n} x \cdot \frac{n!}{(x - 1)! (n - x)!} p^x (1 - p)^{n-x} \] since \( x! = x(x - 1)! \)

    \[ = \sum_{x=1}^{n} \frac{n!}{(x - 1)! (n - x)!} p^x (1 - p)^{n-x} \] cancelling the \( x \)

    This expression is very similar to the probability function for a binomial random variable, and we know the probabilities sum to 1. Taking out factors of \( n \) and \( p \) from the expression and letting \( z = x - 1 \) gives

    \[ E(X) = np \sum_{x=1}^{n} \frac{(n - 1)!}{(x - 1)! (n - x)!} p^{x-1} (1 - p)^{n-x} = np \sum_{z=0}^{n-1} \frac{(n - 1)!}{z! (n - 1 - z)!} p^z (1 - p)^{n-1-z} \]

    Note that this sum corresponds to the sum of all the values of the probability function for a binomial random variable \( Z \), which is the number of successes in \( n - 1 \) trials each with probability of success \( p \). Therefore the sum equals 1, and so

    \[ E(X) = np \]

    Variance

    If \( X \) is a binomial random variable with parameters \( n \) and \( p \), then \( \text{Var}(X) = np(1 - p) \).

    Proof

    The variance of the binomial random variable \( X \) may be found using \( \text{Var}(X) = E(X^2) - \mu^2 \), where \( \mu = np \).

    Thus, to find the variance, we need to determine \( E(X^2) \):

    \[ E(X^2) = \sum_{x=0}^{n} x^2 \cdot \frac{n!}{x!(n-x)!} p^x (1 - p)^{n-x} \]

    But \( x^2 \) is not a factor of \( x! \) and so we cannot proceed as in the previous proof for the expected value.

    The strategy used here is to determine \( E[X(X - 1)] \):

    \[ E[X(X - 1)] = \sum_{x=0}^{n} x(x - 1) \cdot \frac{n!}{x!(n-x)!} p^x (1 - p)^{n-x} \]

    Now the sum corresponds to the sum of all the values of the probability function for a binomial random variable \( Z \), which is the number of successes in \( n - 2 \) trials each with probability of success \( p \), and is thus equal to 1. Hence

    \[ E[X(X - 1)] = n(n - 1)p^2 \]

    \[ E(X^2) - E(X) = n(n - 1)p^2 \]

    Therefore,

    \[ E(X^2) = n(n - 1)p^2 + E(X) \]

    \[ = n(n - 1)p^2 + np \]

    This is an expression for \( E(X^2) \) in terms of \( n \) and \( p \), as required. Thus

    \[ \text{Var}(X) = E(X^2) - \mu^2 \]

    \[ = n(n - 1)p^2 + np - (np)^2 \]

    \[ = np(1 - p) \]

    Example 1

    Finding the probability that of five prisoners chosen at random

    The probability that a person currently in prison has ever been imprisoned before is \(0.72\). We want to find the probability that of five prisoners chosen at random, at least three have been imprisoned before, correct to four decimal places.

    Solution

    If X is the number of prisoners who have been imprisoned before, then

    Pr(X = x) = \(\frac{5!}{x!(5-x)!}(0.72)^x(0.28)^{5-x}\), for \(x = 0, 1, \ldots, 5\)

    So,

    Pr(X ≥ 3) = Pr(X = 3) + Pr(X = 4) + Pr(X = 5)

    = \(\frac{5!}{3!(5-3)!}(0.72)^3(0.28)^2 + \frac{5!}{4!(5-4)!}(0.72)^4(0.28)^1 + \frac{5!}{5!(5-5)!}(0.72)^5(0.28)^0 = 0.8624\)

    Example 2

    Binomial Distribution Probabilities

    For a binomial distribution with \( n = 5 \) and \( p = 0.4 \), find the probability of:

    1. a) No successes

    2. b) Three successes

    3. c) Five successes


    Solution

  • a) No successes:

    Use the formula \( P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \) where \( k = 0 \).

    Plug in the values: \( P(X = 0) = \binom{5}{0} \cdot (0.4)^0 \cdot (1-0.4)^{5-0} = 0.07776 \).

    So, the probability of no successes is \( 0.07776 \).

  • b) Three successes:

    Use the formula \( P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \) where \( k = 3 \).

    Plug in the values: \( P(X = 3) = \binom{5}{3} \cdot (0.4)^3 \cdot (1-0.4)^{5-3} = 0.2304 \).

    So, the probability of three successes is \( 0.2304 \).


  • c) Five successes:

    Use the formula \( P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \) where \( k = 5 \).

    Plug in the values: \( P(X = 5) = \binom{5}{5} \cdot (0.4)^5 \cdot (1-0.4)^{5-5} = 0.01024 \).

    So, the probability of five successes is \( 0.01024 \).


    • Example 3

    Probability of Scoring Goals in Netball

    The probability of a netballer scoring a goal is \(0.3\). We want to find the probability that out of six attempts the netballer scores a goal:

    1. Four times
    2. Four times, given that she scores at least one goal.

    Solution

    Let \( X \) be the number of goals scored. Then \( X \) has a binomial distribution with \( n = 6 \) and \( p = 0.3 \).

    Pr(\( X = 4 \)) = \( \binom{6}{4} \times (0.3)^4 \times (0.7)^2 = 15 \times 0.0081 \times 0.49 = 0.059535 \)

    Pr(\( X = 4 | X \geq 1 \)) = \( \frac{{\text{Pr}(X = 4 \cap X \geq 1)}}{{\text{Pr}(X \geq 1)}} = \frac{{\text{Pr}(X = 4)}}{{\text{Pr}(X \geq 1)}} = \frac{{0.059535}}{{1 - 0.76}} = 0.0675 \) (since \( \text{Pr}(X \geq 1) = 1 - \text{Pr}(X = 0) \))

    Example 4

    The probability of contracting influenza this winter is known to be \[0.2\]. Of the \[100\] employees at a certain business, how many would the owner expect to get influenza?

    Find the standard deviation of the number who will get influenza and calculate \( \mu \pm 2\sigma \).

    Interpret the interval [\( \mu - 2\sigma \), \( \mu + 2\sigma \)] for this example.

    Solution:

    The number of employees who get influenza is a binomial random variable, \( X \), with parameters \( n = 100 \) and \( p = 0.2 \).

    The owner will expect \( \mu = np = 20 \) of the employees to contract influenza.

    The variance is \( \sigma^2 = np(1 - p) = 100 \times 0.2 \times 0.8 = 16 \).

    Hence the standard deviation is \( \sigma = \sqrt{16} = 4 \).

    Thus \( \mu \pm 2\sigma = 20 \pm (2 \times 4) = 20 \pm 8 \).

    The owner of the business knows there is a probability of about 0.95 that from 12 to 28 of the employees will contract influenza this winter.

    Exercise &&1&& (&&1&& Question)

    Suppose that a netball player has a probability of \(\frac{1}{3}\) of scoring a goal each time she attempts to goal. She repeatedly has shots for goal. Is this a Bernoulli sequence?

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    Exercise &&2&& (&&1&& Question)

    Find the probability of obtaining exactly three heads when a fair coin is tossed seven times, correct to four decimal places.

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    Exercise &&3&& (&&1&& Question)

    An examination consists of 30 multiple-choice questions, each question having three possible answers. A student guesses the answer to every question. Let \( X \) be the number of correct answers.

    a) How many will she expect to get correct? That is, find \( E(X) = \mu \).

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    Exercise &&4&& (&&1&& Question)

    b) Find \( \text{Var}(X) \), using the previous question statement.

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