Understanding the Binomial Distribution Problem

The question asks us to find the probability of getting exactly 3 successes out of 5 trials in a Binomial distribution. We are given a key relationship: the mean of this Binomial distribution is three times its variance.

Let's recall the parameters and properties of a Binomial distribution:

• $n$: Number of trials
• $p$: Probability of success in a single trial
• Mean (\(\mu\)): $np$
• Variance (\(\sigma^2\)): $np(1-p)$

In this specific problem, we are given:

• Number of trials, $n = 5$
• Mean = 3 × Variance

We need to find the probability of exactly $k=3$ successes, which is given by the Binomial probability formula:

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

To use this formula, we first need to determine the value of $p$, the probability of success.

Finding the Probability of Success (p)

We can use the given relationship between the mean and variance to find $p$.

Given: \( \text{Mean} = 3 \times \text{Variance} \)

Substituting the formulas for mean and variance:

\( np = 3 \times np(1-p) \)

Since $n=5$ (a non-zero number of trials) and $p$ must be non-zero for a non-degenerate Binomial distribution where mean and variance exist and are positive, we can divide both sides of the equation by $np$ (assuming \(np \neq 0\)).

\( 1 = 3(1-p) \)

Now, we solve for $p$:

\( 1 = 3 - 3p \)

\( 3p = 3 - 1 \)

\( 3p = 2 \)

\( p = \frac{2}{3} \)

So, the probability of success in a single trial is \( \frac{2}{3} \). The probability of failure is \( 1-p = 1 - \frac{2}{3} = \frac{1}{3} \).

Calculating the Probability of Exactly 3 Successes

Now that we have $n=5$, $k=3$, \(p=\frac{2}{3}\), and \(1-p=\frac{1}{3}\), we can use the Binomial probability formula to find the probability of exactly 3 successes:

\( P(X=3) = \binom{5}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^{5-3} \)

\( P(X=3) = \binom{5}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 \)

First, calculate the binomial coefficient \( \binom{5}{3} \):

\( \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10 \)

Next, calculate the powers of the probabilities:

\( \left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27} \)

\( \left(\frac{1}{3}\right)^2 = \frac{1^2}{3^2} = \frac{1}{9} \)

Now, substitute these values back into the probability formula:

\( P(X=3) = 10 \times \frac{8}{27} \times \frac{1}{9} \)

\( P(X=3) = \frac{10 \times 8 \times 1}{27 \times 9} \)

\( P(X=3) = \frac{80}{243} \)

Thus, the probability of exactly 3 successes out of 5 trials is \( \frac{80}{243} \).

Summary of Calculation Steps

Revision Table: Key Binomial Concepts

Additional Information: Properties of Binomial Distribution

The Binomial distribution is based on a series of Bernoulli trials. A Bernoulli trial is a single experiment with only two possible outcomes: success or failure.

Conditions for a process to follow a Binomial distribution:

• There is a fixed number of trials ($n$).
• Each trial is independent of the others.
• There are only two possible outcomes for each trial (success or failure).
• The probability of success ($p$) is the same for every trial.

In this problem, we used the relationship between mean and variance, \( np = 3 \times np(1-p) \), which is a common way to solve for the parameter $p$ if $n$ is known or vice-versa, based on given statistics of the distribution.

The binomial coefficient \( \binom{n}{k} \) represents the number of ways to choose exactly $k$ successes in $n$ trials, without regard to the order.