Problem 8 - 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 one failure

Step-by-Step Solution

Answer

The probability of exactly one failure is approximately 0.36015.

Step 1: Understand the Scenario

We are dealing with a binomial experiment, where we have 5 independent trials. The probability of success (p) is 0.7 and the probability of failure (q) is 0.3.

Step 2: Define the Probability Formula

The probability of achieving exactly k successes in n trials for a binomial distribution is given by the formula: \[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]Where \( \binom{n}{k} \) is the binomial coefficient calculated as \( \frac{n!}{k!(n-k)!} \).

Step 3: Identify the Desired Outcome

We want exactly one failure, which means there will be 4 successes (as there are 5 trials in total). Thus, we need to find \(P(X = 4)\).

Step 4: Calculate the Binomial Coefficient

The binomial coefficient for 4 successes out of 5 trials is:\[ \binom{5}{4} = \frac{5!}{4!1!} = 5 \]

Step 5: Compute the Probability

Using the binomial formula:\[ P(X = 4) = \binom{5}{4} (0.7)^4 (0.3)^{1} \]Calculate each part:- \((0.7)^4 = 0.2401\)- \((0.3)^1 = 0.3\)Thus, \[ P(X = 4) = 5 \times 0.2401 \times 0.3 \approx 0.36015 \]

Step 6: State the Final Probability

The probability of having exactly one failure (or 4 successes) in these 5 trials is approximately 0.36015.

Key Concepts

Probability of Success

In a binomial distribution, the probability of success, often denoted as \( p \), plays a central role. It refers to the probability that a single trial will result in a success.
For our exercise, the probability of success is given as 0.7. This means that there is a 70% chance of success on each individual trial. Since this is a binomial experiment, each trial is independent, and this probability remains constant throughout all trials.
Understanding this constant probability helps us calculate the likelihood of a certain number of successes in a sequence of trials. For instance, if you were flipping a coin with a known probability of landing heads as 0.7, you could predict how often you might achieve a certain number of heads in several flips.
Thus, in scenarios like testing multiple items for quality control or predicting election results, knowing the probability of success is crucial for accurate predictions.

Binomial Coefficient

The binomial coefficient, which is expressed as \( \binom{n}{k} \), is a significant component in calculating binomial probabilities.
The formula for the binomial coefficient is:

• \[ \binom{n}{k} = \frac{n!}{k! (n-k)!} \]

This formula calculates the number of ways to choose \( k \) successes out of \( n \) trials. In our exercise, we calculated \( \binom{5}{4} \), which represents the number of ways to achieve 4 successes out of 5 trials.
In this specific case:

• \( n = 5 \) (total trials)
• \( k = 4 \) (successes needed)
• Thus, \( \binom{5}{4} = 5 \)

The binomial coefficient tells us there are 5 possible ways to arrange four successful trials with one failure.
Understanding the binomial coefficient helps in situations where order and selection count. Whether for probabilities or arranging items, this foundational concept reveals the possible combinations efficiently.

Independent Trials

Independent trials mean that the outcome of any given trial does not affect the outcome of another. In a binomial distribution, this independence is crucial for maintaining factor simplicity.
For example, in our exercise, we have 5 independent trials, where each trial follows the same rules:

• The probability of success, \( p = 0.7 \)
• The probability of failure, \( q = 0.3 \)

Since each trial is independent, the outcome of one does not change the probabilities of the subsequent trials. Whether the previous trial was a success or a failure, the probabilities remain \( p = 0.7 \) and \( q = 0.3 \) in the next trial.
This concept of independence allows us to multiply probabilities over several trials, enabling predictions over sequences of trials. It's like rolling a fair die; each roll is not influenced by what happened before or after. This makes computations straightforward when deriving overall probabilities through repeated events.