Problem 67 - Algebra 2

Question

Open-Ended Write a probability problem for which \(_{5} \mathrm{C}_{3}(0.5)^{2}(0.5)^{3}\) is the solution.

Step-by-Step Solution

Answer

A problem that matches the given expression could be: 'Suppose you flip a coin 5 times. What is the probability of getting exactly 3 heads (assuming heads is a 'success') and 2 tails (assuming tails is a 'failure')?' The solution to this problem is \(_5C_3(0.5)^2(0.5)^3\), which matches the given expression.

Step 1: Identify the total number of trials

We see from the combination \(_5C_3\), the total number of trials equals to 5 because the number \(_5C_3\) refers to the number of ways to choose 3 outcomes from 5 trials.

Step 2: Identify the number of 'success' trials

In our constructed problem, three 'successes' will occur. The binomial coefficient \(_5C_3\) refers to the number of ways to pick 3 successes from 5 trials.

Step 3: Identify the binomial probability of 'success' and 'failure'

The problem conditions must indicate that each trial is independent, with a 50% chance of 'success' and 50% chance of 'failure'. These probabilities are represented by \((0.5)^2\) for the 'successes' and \((0.5)^3\) for 'failures'

Step 4: Construct the Problem

Based on the previous steps, a potential problem could be: 'Suppose you flip a coin 5 times. What is the probability of getting exactly 3 heads (assuming heads is a 'success') and 2 tails (assuming tails is a 'failure')?' The solution to this problem would be \(_5C_3(0.5)^2(0.5)^3\).

Key Concepts

Combinations

In probability theory, combinations help in determining the number of ways to select a subset of items from a larger set, irrespective of the order. This is an essential concept when dealing with binomial probability problems.
For example, if you have 5 trials and you want to find out how many ways you can choose 3 successful outcomes, you use combinations. The mathematical notation used is \( \binom{n}{k} \), where \( n \) is the total number of items to choose from and \( k \) is the number of items to be chosen.
In our example, the combination would be \( _5C_3 \), which calculates the number of different ways to achieve 3 successes out of 5 trials. It is calculated as:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = 10 \]
This tells us there are 10 possible ways you can get exactly 3 successes in 5 trials.

Independent Trials

Independent trials are an essential concept to understand when tackling binomial probability problems. Basically, in such trials, the outcome of one trial does not affect the outcome of another. This is a key aspect of the classic binomial model often used in scenarios like coin flips or quality control tests.

In our illustrative problem, each coin flip is an independent trial because getting heads or tails in one flip does not influence the result of the next flip.

• This independence ensures that the probability of success remains constant throughout each trial.
• For the problem at hand, this means each coin flip has a 50% probability of being a 'success' (heads) or a 'failure' (tails).

Understanding independence is crucial because it ensures that the calculations remain consistent across the duration of the experiment.

Success and Failure Probabilities

In a binomial probability scenario, it's crucial to define the probabilities of success and failure for each trial. These probabilities remain constant due to the assumption of independent trials.
Let's break it down:

In our exercise of flipping a coin, the probability of success (in this context, flipping heads) is 0.5, or 50%, as is the probability of failure (flipping tails). This is a simple yet foundational example for understanding binomial probability.

• The formula \((0.5)^2 \cdot (0.5)^3\) in our exercise showcases the 50% probability for every trial outcome.
• The \((0.5)^2\) refers to the probability of two successes, while \((0.5)^3\) caters to three failures.

When you multiply these probabilities by the combination count, you get the overall probability of getting the desired number of successes and failures. This constant probability framework allows us to precisely compute outcomes in binomial probability problems.