Understanding the Binomial Distribution Problem

The question involves a random variable X that follows a binomial distribution, denoted as B(n, p). Here, n represents the number of trials, and p represents the probability of success in a single trial. We are given that X follows B(6, p), which means the number of trials is n = 6. We are also given a condition relating the probabilities of specific outcomes: \(16{\rm{\;P\;}}\left( {{\rm{X}} = 4} \right) = {\rm{P\;}}\left( {{\rm{X}} = 2} \right)\). Our goal is to find the value of p.

Applying the Binomial Probability Formula

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

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

where:

• n is the number of trials.
• k is the number of successful outcomes we are interested in.
• p is the probability of success on a single trial.
• \((1-p)\) is the probability of failure on a single trial.
• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient, representing the number of ways to choose k successes in n trials.

For our problem, n = 6. We need to find the expressions for \(P(X=4)\) and \(P(X=2)\).

For \(P(X=4)\):

• k = 4
• \(P(X=4) = \binom{6}{4} p^4 (1-p)^{6-4} = \binom{6}{4} p^4 (1-p)^2\)
• First, calculate the binomial coefficient \(\binom{6}{4}\):
• \(\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(2 \times 1)} = \frac{6 \times 5}{2 \times 1} = 15\)
• So, \(P(X=4) = 15 p^4 (1-p)^2\)

For \(P(X=2)\):

• k = 2
• \(P(X=2) = \binom{6}{2} p^2 (1-p)^{6-2} = \binom{6}{2} p^2 (1-p)^4\)
• Next, calculate the binomial coefficient \(\binom{6}{2}\):
• \(\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(4 \times 3 \times 2 \times 1)} = \frac{6 \times 5}{2 \times 1} = 15\)
• So, \(P(X=2) = 15 p^2 (1-p)^4\)

Solving the Probability Equation

We are given the condition: \(16{\rm{\;P\;}}\left( {{\rm{X}} = 4} \right) = {\rm{P\;}}\left( {{\rm{X}} = 2} \right)\). Substitute the expressions we found for \(P(X=4)\) and \(P(X=2)\) into this equation:

\(16 \times \left(15 p^4 (1-p)^2\right) = 15 p^2 (1-p)^4\)

Now, let's solve for p. We can divide both sides by 15 (assuming \(15 \neq 0\), which is true):

\(16 p^4 (1-p)^2 = p^2 (1-p)^4\)

Move all terms to one side to set the equation to zero:

\(16 p^4 (1-p)^2 - p^2 (1-p)^4 = 0\)

Factor out common terms. Both terms have \(p^2\) and \((1-p)^2\):

\(p^2 (1-p)^2 [16 p^2 - (1-p)^2] = 0\)

This equation holds if any of the factors are zero:

1. \(p^2 = 0 \implies p = 0\). This is a valid probability.
2. \((1-p)^2 = 0 \implies 1-p = 0 \implies p = 1\). This is a valid probability.
3. \(16 p^2 - (1-p)^2 = 0\). This is a difference of squares, \((4p)^2 - (1-p)^2\).

Let's solve the third case:

\((4p)^2 - (1-p)^2 = 0\)

Using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\):

\((4p - (1-p))(4p + (1-p)) = 0\)

\((4p - 1 + p)(4p + 1 - p) = 0\)

\((5p - 1)(3p + 1) = 0\)

This gives two possibilities:

• \(5p - 1 = 0 \implies 5p = 1 \implies p = \frac{1}{5}\). This is a valid probability since \(0 \le \frac{1}{5} \le 1\).
• \(3p + 1 = 0 \implies 3p = -1 \implies p = -\frac{1}{3}\). This is not a valid probability since probability must be non-negative.

The possible values for p that satisfy the equation are 0, 1, and \( \frac{1}{5} \). A probability 'p' in a binomial distribution must be a value between 0 and 1, inclusive (\(0 \le p \le 1\)). All three values (0, 1, \( \frac{1}{5} \)) are mathematically valid probabilities.

Comparing these valid solutions with the given options, we find that \( \frac{1}{5} \) is present among the options.

The options were:

• \( \frac{1}{3} \)
• \( \frac{1}{4} \)
• \( \frac{1}{5} \)
• \( \frac{1}{6} \)

The value \( \frac{1}{5} \) matches one of the provided options.

Final Calculation Steps

Let's summarize the calculation:

1. Start with the given equation: \(16 P(X=4) = P(X=2)\) for \(X \sim B(6, p)\).
2. Write \(P(X=k) = \binom{6}{k} p^k (1-p)^{6-k}\).
3. Calculate \(P(X=4) = \binom{6}{4} p^4 (1-p)^2 = 15 p^4 (1-p)^2\).
4. Calculate \(P(X=2) = \binom{6}{2} p^2 (1-p)^4 = 15 p^2 (1-p)^4\).
5. Substitute into the equation: \(16 \times 15 p^4 (1-p)^2 = 15 p^2 (1-p)^4\).
6. Simplify: \(16 p^4 (1-p)^2 = p^2 (1-p)^4\).
7. Rearrange: \(p^2 (1-p)^2 [16 p^2 - (1-p)^2] = 0\).
8. Solve \(16 p^2 - (1-p)^2 = 0 \implies (4p - (1-p))(4p + (1-p)) = 0 \implies (5p-1)(3p+1)=0\).
9. Solutions for p are \(p=0\), \(p=1\), \(p=\frac{1}{5}\), \(p=-\frac{1}{3}\).
10. Valid probabilities are \(0 \le p \le 1\). Thus, p can be 0, 1, or \( \frac{1}{5} \).
11. From the options, \( \frac{1}{5} \) is a possible value for p.

Revision Table: Key Concepts

Additional Information: Properties of Binomial Distribution

The binomial distribution is a fundamental discrete probability distribution. Understanding its properties is crucial for solving related problems.

• Mean: The expected number of successes in n trials is given by \(E(X) = np\).
• Variance: The variance of the number of successes is given by \(Var(X) = np(1-p)\).
• Standard Deviation: The standard deviation is the square root of the variance, \(SD(X) = \sqrt{np(1-p)}\).
• Conditions for Binomial Distribution:
  • There are a fixed number of trials (n).
  • Each trial has only two possible outcomes (success or failure).
  • The probability of success (p) is the same for each trial.
  • The trials are independent of each other.

In this problem, we used the PMF to set up an equation based on the given condition and solved for the unknown parameter p. The solution involved algebraic manipulation and understanding the constraints on the value of a probability.