Problem 43 - A Graphical Approach to Precalculus with Limits

Question

A die is rolled 12 times. Approximate the probability of rolling the following. No more than 3 ones

Step-by-Step Solution

Answer

The probability of rolling no more than 3 ones is approximately 0.589.

Step 1: Identify the probability model

Rolling a die 12 times with a focus on the number of `1s` rolled follows a Binomial distribution. The number of trials, \(n\), is 12, and the probability of success (rolling a one), \(p\), is \(\frac{1}{6}\) since a die has six sides.

Step 2: Define the event of interest

The event we are interested in is rolling no more than 3 ones. This means the probability of rolling 0, 1, 2, or 3 ones. We are looking for \(P(X \leq 3)\).

Step 3: Use the Binomial probability formula

The probability of rolling exactly \(k\) ones in 12 rolls is given by the formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \(\binom{n}{k}\) is the binomial coefficient "n choose k".

Step 4: Calculate probabilities for k = 0, 1, 2, and 3

Compute each probability: - \(P(X=0) = \binom{12}{0}\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^{12}\)- \(P(X=1) = \binom{12}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^{11}\)- \(P(X=2) = \binom{12}{2}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{10}\)- \(P(X=3) = \binom{12}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^{9}\)Calculate each using a calculator for the binomial coefficients and powered terms.

Step 5: Sum the probabilities

Add together the probabilities from Step 4 to find \(P(X \leq 3)\): \[ P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \].

Key Concepts

Probability models

Probability models help us think about and calculate the likelihood of different outcomes in uncertain situations. In the case of rolling a die 12 times, we use something called a Binomial distribution to predict how many times a particular result, such as rolling a 1, will occur. The reason behind choosing a Binomial distribution is quite straightforward: we have a fixed number of independent trials (12 die rolls in this case), and each trial has two outcomes — rolling a one (success) or not rolling a one (failure). The number of successes in these 12 independent trials follows the Binomial probability model. This model is characterized by two key parameters, \(n\) and \(p\). Here, \(n\) is the number of trials, which is 12, and \(p\) is the probability of success in a single trial, which is \(\frac{1}{6}\) since a die has six sides. Probability models like this one allow us to make sense of real-world problems that involve uncertainty.

Binomial probability formula

The Binomial probability formula is the main tool we use to calculate the likelihood of a specific number of successes in a series of independent trials. This formula is written as: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:

• \(P(X = k)\) is the probability of getting exactly \(k\) successes.
• \(\binom{n}{k}\) is the binomial coefficient, often read as "\(n\) choose \(k\)", which tells us how many ways \(k\) successes can occur in \(n\) trials.
• \(p\) is the probability of success on a single trial.
• \((1-p)\) is the probability of failure on a single trial.
• \(n\) is the total number of trials.
• \(k\) is the number of successes we're interested in.

To solve the problem, we apply this formula for \(k = 0, 1, 2,\) and \(3\), which means we calculate how likely it is to roll either 0, 1, 2, or 3 ones during the 12 rolls. Each calculation uses the formula's components to give us the probabilities for these specific outcomes.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting combinations and permutations of different elements. In our exercise, combinatorics is crucial because it provides us with the binomial coefficient \(\binom{n}{k}\), which is an essential part of the Binomial probability formula. The binomial coefficient \(\binom{n}{k}\) denotes the number of ways to choose \(k\) successes from \(n\) trials without considering the order of success. Calculating \(\binom{n}{k}\) involves using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \(!\) denotes a factorial, which is the product of all positive integers up to that number. For example, to find the probability of rolling exactly three ones, we compute \(\binom{12}{3}\), which tells us how many ways three ones can occur across 12 rolls. Combinatorics helps us by providing the foundation to understand how different outcomes relate to each other and how they combine, which is vital for using probability models accurately.