Problem 22 - Algebra 2

Question

As a maintenance manager, Jackie Thomas is responsible for managing the maintenance of an office building. When entering a room after hours, the probability that she selects the correct key on the first try is \(\frac{1}{5} .\) If she enters 6 rooms in an evening, find each probability. \(P(\text { correct exactly } 2 \text { times })\)

Step-by-Step Solution

Answer

The probability of selecting the correct key exactly 2 times out of 6 is \( \frac{3840}{15625} \).

Step 1: Identify the Type of Probability Problem

This problem falls under the category of a binomial probability distribution, where we have a fixed number of trials, two outcomes (correct or incorrect), and the probability of success is constant for each trial.

Step 2: Define the Terms of the Binomial Distribution

We have the probability of success (selecting the correct key), \( p = \frac{1}{5} \), and the probability of failure, \( q = 1 - p = \frac{4}{5} \). The number of trials, \( n \), is 6 rooms. We are asked to find the probability of exactly 2 successes, \( k = 2 \).

Step 3: Use the Binomial Probability Formula

The binomial probability formula is given by \( P(X = k) = \binom{n}{k} p^k q^{n-k} \). We will use this formula to find \( P(X = 2) \).

Step 4: Calculate the Binomial Coefficient

Compute the binomial coefficient for \( \binom{6}{2} \) using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). This results in \( \binom{6}{2} = \frac{6!}{2!4!} = 15 \).

Step 5: Compute the Probability Component

Calculate the probability component: \( p^k = (\frac{1}{5})^2 = \frac{1}{25} \) and \( q^{n-k} = (\frac{4}{5})^4 = \frac{256}{625} \).

Step 6: Calculate the Final Probability

Combine the results from the previous steps to get \( P(X = 2) = 15 \cdot \frac{1}{25} \cdot \frac{256}{625} = \frac{3840}{15625} \).

Step 7: Simplify the Probability Fraction

Simplify \( \frac{3840}{15625} \) if possible. The fraction is already in simplest form.

Key Concepts

Probability of Success

In the context of binomial probability distribution, the probability of success refers to the chance of a particular desired outcome occurring in a single trial. Essentially, it answers the question: "What are the chances of achieving what we want in one attempt?"

In our example, the probability of success is selecting the correct key to open a door. This is represented as \( p \) and has a value of \( \frac{1}{5} \), meaning there is a 20% chance that Jackie picks the correct key on the first try.

This probability remains the same for each of the 6 trials (in this case, rooms), as each selection is an independent event, a key concept in binomial probability. Knowing the probability of success sets the stage for calculating other probabilities in the scenario.

Binomial Coefficient

The binomial coefficient is a central element in the calculation of binomial probabilities. It represents the number of ways to choose a certain number of successes from a set of trials. Mathematically, this is shown as \( \binom{n}{k} \), where \( n \) is the total number of trials and \( k \) is the number of successes we are interested in.

In our problem, where Jackie wants to open 6 doors and is interested in how many ways she can succeed exactly 2 times, the binomial coefficient is calculated as \( \binom{6}{2} \). Using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), we find that there are 15 possible combinations. The factorial function (!) multiplies all whole numbers from its chosen number down to one, helping to calculate combinations effectively.

Binomial Probability Formula

The binomial probability formula is the backbone of determining the likelihood of a specific number of successes in a binomial distribution. This formula is expressed as:

\[ P(X = k) = \binom{n}{k} p^k q^{n-k} \]

This formula combines the binomial coefficient, the probability of success raised to the power of the number of successes, and the probability of failure raised to the number of failures. In our maintenance scenario, it helps us calculate the probability that Jackie will select the correct key exactly twice in six attempts.

After calculating the coefficient as 15, the probability of success as \( (\frac{1}{5})^2 \), and the probability of failure as \( (\frac{4}{5})^4 \), we multiply these elements:

\[ P(X = 2) = 15 \cdot \frac{1}{25} \cdot \frac{256}{625} = \frac{3840}{15625} \].

This gives us the final probability for the desired outcome.

Probability of Failure

In any binomial probability setting, the probability of failure complements the probability of success. This is often denoted as \( q \), where \( q = 1 - p \). It indicates the likelihood of not achieving the desired outcome in a trial.

For Jackie's task with keys, \( q \) is equal to \( \frac{4}{5} \) or 80%, which represents the chance that she does not pick the right key in one attempt.

Understanding the probability of failure is important because it is necessary for computing the \( q^{n-k} \) term in the binomial probability formula. It ensures that all possibilities are accounted for when calculating the chance of exactly \( k \) successes. By multiplying the powers of the failure probability with the success probability, we accurately capture the nature of random events over multiple trials.