To find \(P(2k)\), substitute \(t = 2k\) into the formula:

["# How to Find ( P(2k) ): Substituting ( t = 2k ) into the Formula", "When working with probability distributions—especially binomial distributions—Calculating the probability ( P(X = 2k) ), the event that a random variable ( X ) takes an even value like ( 2k ), often involves substituting variables strategically. One powerful technique is substituting ( t = 2k ) into the binomial probability formula. This substitution simplifies the expression and makes evaluating ( P(2k) ) more straightforward.", "## Understanding the Binomial Distribution", "The binomial probability formula gives the probability of obtaining exactly ( k ) successes in ( n ) independent Bernoulli trials:", "[\nP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}\n]", "Here,\n- ( n ) is the total number of trials,\n- ( k ) is the number of successful trials,\n- ( p ) is the probability of success on a single trial,\n- ( \binom{n}{k} ) is the binomial coefficient.", "Sometimes, we want ( P(X = 2k) )—the probability that the number of successes is exactly twice ( k ). Substituting ( t = 2k ), we shift focus to evaluating probabilities at even integers, useful in modeling scenarios like even-numbered wins, sampling with parity, or restricted outcomes.", "## Substituting ( t = 2k ): The Key Step", "To find ( P(2k) ), redefine the number of successes as ( t ). Then:", "[\nP(X = t) = \binom{n}{t} p^t (1 - p)^{n - t}\n]", "Substituting ( t = 2k ), we obtain:", "[\nP(2k) = \binom{n}{2k} p^{2k} (1 - p)^{n - 2k}\n]", "This substitution reformulates the problem in terms of even outcomes, allowing us to:", "- Evaluate probabilities at precisely the values of interest,\n- Simplify expressions involving even powers and binomial coefficients,\n- Apply conditional or recursive methods tailored for even-shaped outcomes.", "## Why This Substitution Matters", "Using ( t = 2k ) transforms the target probability ( P(2k) ) into a standard binomial expression evaluated at a simpler, constrained point. This technique is especially useful in:", "- Statistical modeling, where even outcomes arise naturally (e.g., defective/lchts/faulty items in pairs),\n- Recursive algorithms that compute cumulative probabilities on even indices,\n- Proofs and derivations, simplifying terms with even exponents and indices.", "By substituting early, complex summations over scattered even ( k ) values condense into a clean formula involving a single parameter ( t = 2k ).", "## Practical Application Example", "Problem: Suppose a factory produces light bulbs with a 15% defect rate (( p = 0.15 )). What is ( P(X = 4) )?", "Since ( 4 = 2 \ imes 2 ), let ( t = 4 = 2k \Rightarrow k = 2 ).", "Apply the formula:", "[\nP(2k) = P(4) = \binom{n}{4} (0.15)^4 (0.85)^{n - 4}\n]", "Assume ( n = 10 ) trials:", "[\nP(4) = \binom{10}{4} (0.15)^4 (0.85)^6\n]", "Calculate:", "- ( \binom{10}{4} = 210 )\n- ( (0.15)^4 \approx 0.00050625 )\n- ( (0.85)^6 \approx 0.377149 )", "[\nP(4) \approx 210 \ imes 0.00050625 \ imes 0.377149 \approx 0.0401\n]", "Thus, ( P(2k) = P(4) \approx 0.0401 ) or 4.01%.", "## Conclusion", "Substituting ( t = 2k ) into the binomial probability formula is a strategic move to efficiently compute ( P(2k) ). By redefining the index of success in terms of even integers, this method streamlines calculations, enhances clarity, and supports more complex analyses involving only desired even outcomes. Whether in theoretical derivations or applied statistics, mastering this substitution empowers precise probability modeling with simplicity and elegance.", "---", "Keywords: ( P(2k) ), binomial probability, substitution ( t = 2k ), probability of even outcomes, binomial distribution formula, probability calculation, statistical modeling.", "Meta Description:\nLearn how to compute ( P(2k) ) efficiently by substituting ( t = 2k ) into the binomial formula. Simplify even-valued probabilities with clear steps and practical examples."]