Problem 57 - Basic Statistics for Business and Economics with Student

Question

Refer to Exercise \(56,\) but assume there are 10 homes in the Quail Creek area and four of them have a security system. Three homes are selected at random: a. What is the probability all three of the selected homes have a security system? b. What is the probability none of the three selected homes have a security system? c. What is the probability at least one of the selected homes has a security system? d. Did you assume the events to be dependent or independent?

Step-by-Step Solution

Answer

a. \(\frac{1}{30}\); b. \(\frac{1}{6}\); c. \(\frac{5}{6}\); d. Dependent events.

Step 1: Understanding the Scenario

We know that there are 10 homes in total, and 4 of these homes have a security system. We are selecting 3 homes at random. We need to calculate probabilities based on these selections.

Step 1: Probability All Three Have Security System

To find the probability that all three selected homes have a security system, use the formula for probability without replacement:\[ P(\text{all three have security system}) = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} \]Calculate this step-by-step:- First home with security: \(\frac{4}{10}\)- Second home with security: \(\frac{3}{9}\)- Third home with security: \(\frac{2}{8}\)The product gives us the probability:\[ P = \frac{4}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{1}{30} \]

Step 2: Probability None Have Security System

To find the probability that none have a security system, calculate the probability of selecting homes without one.\[ P(\text{none have security system}) = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} \]- First home without security: \(\frac{6}{10}\)- Second home without security: \(\frac{5}{9}\)- Third home without security: \(\frac{4}{8}\)The product gives us the probability:\[ P = \frac{6}{10} \times \frac{5}{9} \times \frac{4}{8} = \frac{1}{6} \]

Step 3: Probability at Least One Has Security System

To find the probability that at least one home has a security system, use the complement rule. The complement is that none have a system. Thus,\[ P(\text{at least one has security}) = 1 - P(\text{none have security system}) \]From Step 2, we have:\[ P(\text{none have security system}) = \frac{1}{6} \]Therefore,\[ P(\text{at least one has security}) = 1 - \frac{1}{6} = \frac{5}{6} \]

Step 4: Determine if Events are Dependent or Independent

The events are dependent because choosing one home affects the probability of choosing the next. Once a home is selected, it's not replaced, affecting the total number of remaining homes and the probability of selecting a home with or without a security system.

Key Concepts

Dependent Events

In probability, dependent events are those where the outcome or occurrence of the first event affects the probability of the subsequent events. In this exercise, choosing a home affects the remaining pool of homes from which additional choices are made. This is because we are selecting without replacing the homes after each pick. Therefore, each new selection is influenced by the results of previous ones, both in terms of the number of homes left to choose from and the number of homes with or without a security system remaining.
To determine if events are dependent, consider if one event has a direct impact on the next.

• If removing one home decreases the total number of homes from 10 to 9 for the subsequent selection, this is a classic indicator of dependent events.
• The probability of selecting a specific type of home, such as those with a security system, changes as the composition of the remaining pool changes.

Understanding dependent events is crucial when calculating probabilities without replacement.

Probability Without Replacement

When dealing with probability problems where items or scenarios are not replaced after each draw or event, we refer to this as 'Probability Without Replacement'. This concept means that each choice directly alters the conditions for the following choices. In the context of our home-selection exercise, after each home is selected, the total number of homes decreases, as does the number of homes with security systems if such homes are chosen.
This calculation method necessitates updating the probability for each subsequent selection. The formula changes with each step:

• For the first selection, the probability is based on the initial mix, e.g., selecting a home with security: \( \frac{4}{10} \).
• The second selection adjusts for the previous selection, affecting both numerator and denominator, e.g., \( \frac{3}{9} \).
• For the third, further adjust: \( \frac{2}{8} \).

Each event depends on the last, reinforcing that these are dependent events, as each pick alters the probabilities.

Complement Rule

The complement rule is a fundamental principle in probability that helps to find the probability of an event by understanding the probability of its complement, or opposite. The complement of an event \( A \) is denoted by \( A' \), meaning "not \( A \)". The rule states that the probability of an event \( A \) occurring is equal to 1 minus the probability of its complement: \[P(A) = 1 - P(A')\]In our exercise, to find the probability that at least one of the selected homes has a security system, it's easier to first calculate the probability none have a security system, and then apply the complement rule.

• Calculate \( P(\text{none have security}) = \frac{1}{6} \).
• Apply the complement rule: \( P(\text{at least one has security}) = 1 - P(\text{none have security}) \).

Thus, the complement rule simplifies calculations, especially when directly calculating complex probabilities is cumbersome.