Problem 1 - Algebra and Trigonometry

Question

Five independent trials of a binomial experiment with probability of success \(p=0.7\) and probability of failure \(q=0.3\) are performed. Find the probability of each event. Exactly two successes

Step-by-Step Solution

Answer

The probability of exactly two successes is approximately 0.1323.

Step 1: Identify the Parameters

In this binomial distribution problem, we need to identify the number of trials and probabilities: the number of trials (n) is 5, probability of success (p) is 0.7, and probability of failure (q) is 0.3.

Step 2: Recall the Binomial Probability Formula

The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient, \( p \) is the probability of success, and \( k \) is the number of successes.

Step 3: Calculate the Binomial Coefficient

Calculate the binomial coefficient \( \binom{5}{2} \):\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

Step 4: Plug Values into Binomial Probability Formula

We want exactly two successes (k = 2):\[ P(X = 2) = \binom{5}{2} \cdot 0.7^2 \cdot 0.3^{5-2} \]Plug the values into the formula:\[ P(X = 2) = 10 \cdot 0.7^2 \cdot 0.3^3 \]

Step 5: Calculate Powers of Probabilities

Calculate the powers: \[ 0.7^2 = 0.49 \]\[ 0.3^3 = 0.027 \]

Key Concepts

Probability of Success

The probability of success in a binomial experiment refers to the chance that a single trial results in a desired outcome. In our case, each trial has a probability of success (\(p\)) of 0.7, which means there is a 70% chance of the event occurring in any given trial. This value is crucial because it directly influences the likelihood of different possible outcomes when multiple trials are performed.

• Probability of success = \(p\)
• Given \(p = 0.7\) for the problem at hand

The probability of success is always between 0 and 1, inclusive. A value of 0 means the event cannot happen, while a value of 1 means the event is certain. In a binomial distribution, \(p\) remains constant during each trial, reflecting a key feature of independent trials.

Binomial Coefficient

The binomial coefficient is an essential component for solving problems involving binomial distributions. It determines the number of ways to choose \(k\) successes from \(n\) trials. Denoted by \(\binom{n}{k}\), it is calculated through the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

• Factorial (!) means multiplying a series of descending natural numbers. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.
• In our specific problem, \(n = 5\) trials and \(k = 2\) successes, so \(\binom{5}{2}\).
• Solving, we get \(\binom{5}{2} = 10\): \(\frac{5 \times 4}{2 \times 1}\).

The binomial coefficient tells us there are 10 different combinations in which 2 successes can occur in 5 trials, a crucial step in finding the probability of these specific outcomes.

Binomial Probability Formula

The binomial probability formula helps us calculate the probability of exactly \(k\) successes in \(n\) independent trials. The formula is:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Here:

• \(\binom{n}{k}\) is the binomial coefficient, calculated earlier.
• \(p\) is the probability of success on an individual trial.
• \((1-p)\) is the probability of failure for an individual trial.

For our example, we need exactly 2 successes out of 5 trials:

• \(p = 0.7\)
• \(1-p = 0.3\)
• \(P(X = 2) = 10 \cdot 0.7^2 \cdot 0.3^3\)

This formula systematically considers all aspects of a binomial trial to calculate the occurrence likelihood of any specific outcome configuration.

Independent Trials

In a binomial distribution, the concept of independent trials is crucial. Each trial is unaffected by the outcome of any other trial. In our example, this means the probability of success (\(p = 0.7\)) is consistent across all five trials.

• No previous trial results influence the next trial's success or failure.
• Independence ensures that whatever the outcome of one trial, it doesn't change probabilities in subsequent trials.

This helps maintain the simplicity of modeling binomial distributions because it implies a stable background where only the fixed probabilities, not past results, dictate possible future outcomes.