Problem 19 - 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. \(P(\text { never the correct key })\)

Step-by-Step Solution

Answer

The probability is \( \frac{4096}{15625} \).

Step 1: Understanding the Event

To find the probability of never selecting the correct key (denoted as event A), we need to identify that the probability of choosing the incorrect key is the complement of choosing the correct key. If the probability of choosing the correct key is \( \frac{1}{5} \), the probability of choosing the incorrect key is \( 1 - \frac{1}{5} = \frac{4}{5} \).

Step 2: Setting Up the Problem

We need to calculate the probability of Jackie never selecting the correct key over 6 rooms, which is akin to selecting the incorrect key all 6 times. Since each room entry is an independent event, we'll raise the probability of choosing the incorrect key, \( \frac{4}{5} \), to the power of the number of rooms, 6.

Step 3: Computing the Probability

The probability, \( P(\text{never the correct key}) \), is calculated as: \[ P(\text{never the correct key}) = \left( \frac{4}{5} \right)^6 \].

Step 4: Calculating the Result

Now compute the value: \[ \left( \frac{4}{5} \right)^6 = \frac{4096}{15625} \]. Therefore, \( P(\text{never the correct key}) = \frac{4096}{15625} \).

Key Concepts

Complementary Events

In probability, complementary events are a foundational concept, crucial for understanding many real-world scenarios. A complementary event consists of all the outcomes that are not part of the original event. Take, for example, flipping a coin. If we define getting heads as the event, then getting tails is the complementary event.
For this exercise, we see the complementary relationship when Jackie chooses keys. The probability of choosing the correct key is \( \frac{1}{5} \). Thus, the probability of not choosing the correct key is its complement, calculated as \( 1 - \frac{1}{5} = \frac{4}{5} \).
This concept allows us to switch our focus from what happens to what doesn’t happen, helping us solve problems by looking at the total probability, which always sums up to 1. In scenarios where direct calculation seems tough, complementary events provide a handy shortcut.

Independent Events

Independent events in probability are those where the outcome of one event doesn’t affect the outcome of another. A classic example is rolling dice. No matter the result of the first roll, the second roll has no impact on it.
In Jackie’s scenario, when entering each room with a key, the choice for one room does not influence any subsequent room. Each room is an independent trial with its own set probability of success and failure. So, when calculating probabilities over multiple rooms, you treat each attempt as a separate and independent event.
This independence is vital because it allows us to multiply the probabilities of sequential events to determine the combined probability. Therefore, understanding independence helps in correctly setting up calculations for sequences of events.

Probability Calculations

Probability calculations involve assigning numerical values to the likelihood of events occurring. In Jackie’s problem, these calculations are key to determining the chance of all events going one particular way.
We started with determining the probability of an incorrect key selection as \( \frac{4}{5} \). Because Jackie enters 6 rooms, and each attempt is independent, we raise this probability to the sixth power. This calculation gives us \( \left( \frac{4}{5} \right)^6 \).
To find the exact probability, we then compute \( \frac{4096}{15625} \). By breaking down such probability calculations into steps—finding individual probabilities and using multiplication for sequences—we conclude a comprehensive assessment of the probability for the event's outcome.