Problem 66 - Algebra 2

Question

For the given probability of success \(p\) on each trial, find the probability of \(x\) successes in \(n\) trials. $$ x=3, n=5, p=0.6 $$

Step-by-Step Solution

Answer

The probability of obtaining 3 successes in 5 trials is 0.216.

Step 1: Identify variables from the problem

From the problem statement, the number of successes \( x \) is 3, the number of trials \( n \) is 5 and the probability of success \( p \) on each trial is 0.6.

Step 2: Substitute the values into the binomial probability formula

The binomial probability \( P(x) \) can be obtained by substituting the given values into the formula: \( P(3) = \binom{5}{3} \cdot (0.6)^3 \cdot (0.4)^{2} \).

Step 3: Calculate the binomial coefficient

Calculate \( \binom{5}{3} \), which is the number of ways to choose 3 successes from 5 trials. This is calculated as \( \frac{5!}{(5 - 3)!3!} = 10 \), where '!' denotes factorial of a number.

Step 4: Substitute the binomial coefficient back into the formula and calculate the final result

Now that we have the binomial coefficient, we can substitute it back into the formula to find the probability. \( P(3) = 10 \cdot (0.6)^3 \cdot (0.4)^{2} = 0.216 \).

Key Concepts

Probability of Success

When dealing with binomial probability, we often talk about the 'probability of success.' This probability is noted as \( p \) and it represents the likelihood that a single trial will result in success.
For example, in the exercise we are considering, the probability of success is \( p = 0.6 \). This means there is a 60% chance of achieving success in one trial.
Understanding this concept is crucial for calculating probabilities when multiple trials are involved. You'll encounter problems requiring you to find the probability of obtaining a specific number of successes across a set number of trials. Each trial is independent, meaning the outcome of one does not affect the others.

• The probability of failure in a single trial is \( 1 - p \), which is 0.4 in this instance.
• The probability of success remains constant across all trials.

This understanding of the probability of success helps in constructing and solving binomial probability problems effectively.

Binomial Coefficient

The binomial coefficient is a key part of the binomial probability formula. It determines how many ways we can choose a certain number of successes from a larger set of trials.
In our case, the binomial coefficient is represented by \( \binom{5}{3} \), meaning we are looking for the number of ways to choose 3 successes out of 5 trials.
This is important because it shows the different possible scenarios that can result in obtaining those 3 successes. The formula for the binomial coefficient is given by: \[ \binom{n}{x} = \frac{n!}{(n-x)!x!} \] Where \( n! \) (n factorial) denotes the product of all positive integers up to \( n \). This helps to arrange and compute possibilities appropriately.

• In simpler terms, the binomial coefficient answers the question: "In how many different ways could this set of successes occur?"
• It is crucial to use it to calculate the accurate probability in a binomial distribution.

Making sense of this concept aids significantly in understanding the setup of your probability calculations.

Factorial Calculation

The idea of factorial calculation is central in calculating the binomial coefficient. A factorial, noted as \( n! \), is the product of all positive integers less than or equal to \( n \).
For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). This calculation helps when dealing with permutations and combinations in probabilities.
It supports the determination of how many ways you can arrange or select objects.

• In the context of the binomial coefficient, factorials simplify how you choose successes from trials.
• For example: to calculate \( \binom{5}{3} \), you compute \( 5! \), \( 2! \), and \( 3! \) and substitute these values into the formula.

Understanding factorial calculations allows you to utilize them effectively in resolving probability problems that involve choosing or arranging items.