Question

Find the value of \(\left(\begin{array}{l}5 \\\ 0\end{array}\right)\left(\frac{1}{4}\right)^{5}+\left(\begin{array}{l}5 \\\ 1\end{array}\right)\left(\frac{1}{4}\right)^{4}\left(\frac{3}{4}\right)+\left(\begin{array}{l}5 \\\ 2\end{array}\right)\left(\frac{1}{4}\right)^{3}\left(\frac{3}{4}\right)^{2}\) \(+\left(\begin{array}{l}5 \\\ 3\end{array}\right)\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{3}+\left(\begin{array}{l}5 \\\ 4\end{array}\right)\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{4}+\left(\begin{array}{l}5 \\\ 5\end{array}\right)\left(\frac{3}{4}\right)^{5}\)

Short answer

The value of the expression is 1.

Step-by-step solution

Identify the binomial coefficients

The given expression involves binomial coefficients, which can be represented as \(\binom{n}{k}\). Here, \(n = 5\) and the terms range from \(k=0\) to \k=5\. List out all the binomial coefficients involved, \(\binom{5}{0}\), \(\binom{5}{1}\), \(\binom{5}{2}\), \(\binom{5}{3}\), \(\binom{5}{4}\) and \(\binom{5}{5}\).

Compute the binomial coefficients

Calculate each binomial coefficient: \(\binom{5}{0} = 1\), \( \binom{5}{1} = 5 \), \( \binom{5}{2} = 10 \), \( \binom{5}{3} = 10\), \( \binom{5}{4} = 5 \), and \( \binom{5}{5} = 1 \).

Simplify each term

Combine the binomial coefficients with the given fractions and simplify each term step-by-step: \(\binom{5}{0}\frac{1}{4}^5\), \( \binom{5}{1}(\frac{1}{4})^4(\frac{3}{4}) \), \( \binom{5}{2}(\frac{1}{4})^3(\frac{3}{4})^2 \), \( \binom{5}{3}(\frac{1}{4})^2(\frac{3}{4})^3 \), \( \binom{5}{4}(\frac{1}{4})(\frac{3}{4})^4 \), \( \binom{5}{5}(\frac{3}{4})^5 \).

Calculate numerical values of each term

Evaluate the simplified expressions to find each term: \[ 1 \times (\frac{1}{4})^5 = \frac{1}{1024} \] \[ 5 \times (\frac{1}{4})^4 \times \frac{3}{4} = \frac{15}{1024} \] \[ 10 \times (\frac{1}{4})^3 \times (\frac{3}{4})^2 = \frac{90}{1024} \] \[ 10 \times (\frac{1}{4})^2 \times (\frac{3}{4})^3 = \frac{270}{1024}\] \[ 5 \times (\frac{1}{4}) \times (\frac{3}{4})^4 = \frac{405}{1024} \] \[ 1 \times (\frac{3}{4})^5 = \frac{243}{1024} \]

Sum all the terms

Add up all the computed values: \(\frac{1}{1024} + \frac{15}{1024} + \frac{90}{1024} + \frac{270}{1024} + \frac{405}{1024} + \frac{243}{1024}\).

Simplify the sum

Combine the fractions: \(\frac{1+15+90+270+405+243}{1024} = \frac{1024}{1024} = 1\).

Key concepts

Binomial Coefficients

The binomial coefficients are a crucial part of the binomial theorem. They are represented as \( \binom{n}{k} \), where \( n\) is the total number of items, and \( k \) is the number of items to be chosen. These coefficients tell you how many ways you can choose \( k \) items from \( n \) items without considering the order. You can calculate binomial coefficients using the formula \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]. Here, \( ! \) represents factorial, which is the product of all positive integers up to that number. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). For the exercise given, the binomial coefficients involved are: \( \binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \binom{5}{3}, \binom{5}{4}, \binom{5}{5} \).

Pascal's Triangle

A simple way to find binomial coefficients is by using Pascal's Triangle. Each number in the triangle is the sum of the two numbers directly above it. The top row is row 0 and starts with the number 1. The next row has two 1s, and so forth. For instance, row 5 in Pascal's Triangle is \[ 1, 5, 10, 10, 5, 1 \]. These numbers correspond to the binomial coefficients we calculated earlier: \( \binom{5}{0} = 1, \binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1 \). This triangle is very helpful for quickly identifying binomial coefficients, especially for small values of \( n \).

Binomial Expansion

The binomial theorem allows you to expand expressions of the form \( (a + b)^n \). This expansion involves summing terms that include binomial coefficients, powers of \( a \), and powers of \( b \). A general term in the binomial expansion can be written as \[ \binom{n}{k} a^{n-k} b^k \]. In the given exercise, the expression is an expansion of \( (\frac{1}{4} + \frac{3}{4})^5 \). Each term in the expansion is represented as \[ \binom{5}{k} (\frac{1}{4})^{5-k} (\frac{3}{4})^k \] for \( k \) ranging from 0 to 5. Expanding this, you get six terms which are combined to form the final expression. Understanding each term's derivation is crucial for mastering binomial expansion.

Probability

Binomial expansions are also useful in probability, especially in binomial distributions. In a binomial distribution, you calculate the probability of getting exactly \( k \) successes in \( n \) independent trials, where the probability of success in each trial is \( p \). The formula used is: \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \]. In the exercise, the terms resemble the form of binomial distribution probabilities, though this exercise is focused on expanding rather than calculating actual probabilities. However, knowing how to express and expand these terms lays the groundwork for calculating probabilities in binomial experiments.