Solving Binomial Distribution Probability for 'p'

This problem involves a Binomial distribution B(n, p) where we are given the number of trials, n = 6, and a relationship between two specific probabilities: 9 times the probability of getting 4 successes is equal to the probability of getting 2 successes, i.e., 9P(X = 4) = P(X = 2).

The probability mass function (PMF) for a Binomial distribution B(n, p) is given by the formula:

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

where:

• \(n\) is the number of trials.
• \(k\) is the number of successful outcomes.
• \(p\) is the probability of success on a single trial.
• \(\binom{n}{k}\) is the binomial coefficient, calculated as \(\frac{n!}{k!(n-k)!}\).

In this problem, we have n = 6. We are given the relationship 9P(X = 4) = P(X = 2).

Let's write out the expressions for P(X = 4) and P(X = 2) using the PMF formula with n = 6:

• For X = 4 (k = 4):

\( P(X = 4) = \binom{6}{4} p^4 (1-p)^{6-4} = \binom{6}{4} p^4 (1-p)^2 \)

• For X = 2 (k = 2):

\( P(X = 2) = \binom{6}{2} p^2 (1-p)^{6-2} = \binom{6}{2} p^2 (1-p)^4 \)

Now, let's calculate the binomial coefficients:

• \(\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{6 \times 5}{2} = 15\)
• \(\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{6 \times 5}{2} = 15\)

Substitute these values back into the probability expressions:

• \( P(X = 4) = 15 p^4 (1-p)^2 \)
• \( P(X = 2) = 15 p^2 (1-p)^4 \)

Now, use the given relationship 9P(X = 4) = P(X = 2):

\( 9 \times [15 p^4 (1-p)^2] = [15 p^2 (1-p)^4] \)

\( 135 p^4 (1-p)^2 = 15 p^2 (1-p)^4 \)

To solve for 'p', we can simplify this equation. We can divide both sides by common terms. Assuming \(p \neq 0\), \(p \neq 1\), we can divide by \(15 p^2 (1-p)^2\).

\( \frac{135 p^4 (1-p)^2}{15 p^2 (1-p)^2} = \frac{15 p^2 (1-p)^4}{15 p^2 (1-p)^2} \)

This simplifies to:

\( \frac{135}{15} \times \frac{p^4}{p^2} \times \frac{(1-p)^2}{(1-p)^2} = \frac{15}{15} \times \frac{p^2}{p^2} \times \frac{(1-p)^4}{(1-p)^2} \)

\( 9 \times p^{4-2} \times 1 = 1 \times 1 \times (1-p)^{4-2} \)

\( 9 p^2 = (1-p)^2 \)

Now, take the square root of both sides:

\( \sqrt{9 p^2} = \pm \sqrt{(1-p)^2} \)

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

Since 'p' is a probability, \(0 \le p \le 1\). This means \(p \ge 0\) and \(1-p \ge 0\). So, \(|p| = p\) and \(|1-p| = 1-p\). The equation becomes:

\( 3p = 1-p \)

Solve for p:

\( 3p + p = 1 \)

\( 4p = 1 \)

\( p = \frac{1}{4} \)

We should also consider the case from taking the negative square root: \(3p = -(1-p)\)

\( 3p = -1 + p \)

\( 3p - p = -1 \)

\( 2p = -1 \)

\( p = -\frac{1}{2} \)

However, the probability 'p' must be between 0 and 1 (inclusive). Since \(p = -\frac{1}{2}\) is outside this range, it is not a valid probability in this context.

Thus, the only valid value for p is \(\frac{1}{4}\).

Revision Table: Key Binomial Concepts

Let's quickly review some essential concepts related to the Binomial distribution.

Additional Information: Properties of Probability

When working with probabilities, it's crucial to remember some fundamental properties. In the context of the Binomial distribution and finding the value of 'p', the probability of success:

• Must be a non-negative value. Probability cannot be negative.
• Must be less than or equal to 1. The probability of an event happening cannot exceed 1 (or 100%).
• Therefore, for any probability 'p', it must satisfy the condition \(0 \le p \le 1\). This was important when we had two possible values for 'p' from the algebraic solution, as only one fell within the valid range for a probability.

Understanding these basic properties helps in interpreting results and validating solutions in probability problems, including those involving the Binomial distribution.