Binomial Distributions

A binomial distribution is a type of Probability Distribution that follows these conditions:

There are multiple Trials
Each Trial has only two possible outcomes: Success or Failure
The Probability of Success is the same in every Trial
The number of Successes in a fixed number of Trials is the Random Variable
The outcome of one Trial does not affect the outcome of others (Independence)

Probability in a Binomial Distribution

The binomial probability formula calculates the probability of getting exactly \( x \) Successes in \( n \) Trials:

\(P(X = x) = \dbinom{n}{x} p^x q^{n-x}\)

Where:

\(P(X = x)\) represents the probability of having exactly \( x \) successes in \( n \) trials
\(\binom{n}{x}\) represents the binomial coefficient, or the number of ways to choose \( x \) Successes out of \( n \) Trials
\(n\) represents the total number of trials or experiments
\(x\) represents the number of successes we are interested in (a value between 0 and \(n\))
\(p\) represents the probability of success on any single trial

Expectation for a Binomial Distribution

The expected number of successes can be determined algebraically as such:

\(E(X) = np\)

Where:

\(E(X)\) represents the Expected Value, or the average number of Successes over many sets of \( n \) Trials
\(n\) represents the total number of Trials or Experiments
\(p\) represents the Probability of Success on any single Trial

Example

Suppose you roll a fair six-sided die 10 times. What is the Expected Number of times you will roll a 3?

Since the die is fair, the Probability of rolling any specific number (like a 3) is:

\(\cfrac{1}{6}\)

Next, we can plug the pertinent values into the Expected Value formula:

\(E(X) = np \)

\(E(X) = 10 \times \cfrac{1}{6}\)

\(E(X) = 1.67\)

Thus, we can determine the Expected Number of times you will roll a 3 is 1.67.

Lee’s family moves to an area with a different telephone exchange. Their new number has four random digits. Lee’s favorite numbers are prime digits: 2, 3, 5, and 7.

Calculate the Probability Distribution for the number of these prime digits
What is the Expected Number of these prime digits?

Show Answer

i. First, note that the prime digits between 0 and 9 are 2, 3, 5, and 7 (not 11 as it's not a single digit). There are 10 possible digits (0-9), so the probability of a digit being prime is 4 out of 10.

The number of prime digits \(X\) follows a binomial distribution: \(X \sim \text{Binomial} \left(n = 4, p = \dfrac{4}{10}\right) \).

The Probability Mass Function for a Binomial Distribution is:

\(P(X = k) = \dbinom{n}{k} p^k (1-p)^{n-k}\)

Where \(n = 4\), \( p = \dfrac{4}{10} = 0.4 \), and \( k \) is the number of prime digits.

Next, we can determine the probability distributions for \(k = 0, 1, 2, 3, 4\):

For \( X = 0 \):

\[ P(0) = \binom{4}{0} (0.4)^0 (0.6)^4 = 1 \times 1 \times 0.1296 = 0.1296 \]

For \( X = 1 \):

\[ P(1) = \binom{4}{1} (0.4)^1 (0.6)^3 = 4 \times 0.4 \times 0.216 = 0.3456 \]

For \( X = 2 \):

\[ P(2) = \binom{4}{2} (0.4)^2 (0.6)^2 = 6 \times 0.16 \times 0.36 = 0.3456 \]

For \( X = 3 \):
\[P(3) = \binom{4}{3} (0.4)^3 (0.6)^1 = 4 \times 0.064 \times 0.6 = 0.1536 \]
For \( X = 4 \):

\[ P(4) = \binom{4}{4} (0.4)^4 (0.6)^0 = 1 \times 0.0256 \times 1 = 0.0256 \]

Therefore, we can determine the probability distributions for \(k = 0, 1, 2, 3, 4\) are \(\mathbf{0.1296, 0.3456, 0.3456, 0.1536,}\) and \(\mathbf{0.0256}\) respectively.

ii. In order to determine the Expected Value, we can plug the pertinent values into the corresponding formula and solve:

\(E(X) = np\)

\(E(X) = 4 \times 0.4\)

\(E(X) = 1.6\)

So, the Expected Number of prime digits is 1.6.