Problem 5 - Algebra 2

Question

There are 8 action, 3 comedy, and 5 children's DVDs on a shelf. Suppose two DVDs are selected at random from the shelf. Find each probability. \(P(\text { a comedy DVD, then a children's DVD), if no replacement occurs }\)

Step-by-Step Solution

Answer

The probability is \( \frac{1}{16} \).

Step 1: Determine Total DVDs

Count the total number of DVDs on the shelf. There are 8 action, 3 comedy, and 5 children's DVDs, which sum up to 16 DVDs in total.

Step 2: Calculate Probability of First Event

Find the probability of selecting a comedy DVD first. There are 3 comedy DVDs, so the probability is given by \( \frac{3}{16} \).

Step 3: Determine Remaining DVDs

After selecting one comedy DVD, there are now 15 DVDs left on the shelf (1 less).

Step 4: Calculate Probability of Second Event

Find the probability of selecting a children's DVD given that a comedy DVD has already been selected and not replaced. There are still 5 children's DVDs left, so the probability is \( \frac{5}{15} \).

Step 5: Calculate Overall Probability

Multiply the probabilities of the two independent events to find the overall probability. This is \( \frac{3}{16} \times \frac{5}{15} = \frac{1}{16} \).

Key Concepts

Combinatorics

Combinatorics is a crucial area of mathematics focusing on counting, arrangement, and combination of objects from a set. In this exercise, combinatorics allows us to calculate probabilities by considering all possible ways to select DVDs.
For instance, when picking two DVDs out of a shelf containing different kinds, we use combinatorial methods to count the total ways of selection. First, find how many DVDs are there altogether, which is needed to set the stage for probability calculations.
This might involve simple addition, like summing the different DVDs on the shelf to find out there are 16 DVDs in total. Without combinatorics, we would not be able to determine the initial set of conditions needed to calculate probability.

Conditional Probability

Conditional probability deals with finding the probability of an event occurring, given another event has already happened. In our exercise, we are tasked to find the probability of picking a children's DVD after already picking a comedy DVD without replacement.
First, calculate the probability of the first event: selecting a comedy DVD. There are 3 comedy DVDs out of a total of 16 DVDs, leading to a probability of \( \frac{3}{16} \).
Once a comedy DVD is picked, it is not replaced. This changes the total number of DVDs on the shelf to 15. The condition now is that we must find the probability of choosing a children's DVD from these remaining DVDs. Since there are still 5 children's DVDs left, our updated probability becomes \( \frac{5}{15} \).
The concept emphasizes that our second probability is dependent on the outcome of the first event—hence the term "conditional probability." To find the combined probability of both events occurring in sequence, we multiply the two probabilities together (\( \frac{3}{16} \times \frac{5}{15} \)).

Independent Events

The term "independent events" typically refers to scenarios where the occurrence of one event does not affect the probability of another. However, in dealing with sequences like our DVD selection, where each step modifies the conditions (like removing one DVD), we often face dependent events.
In this DVD example, though labeled as "independent" in step 5, it more accurately represents conditional dependency because selecting one DVD affects the remaining pool. A genuine case of independent events would be replacing the DVD on the shelf before selecting another; this would keep the total number constant for each pick.
The process in the problem, however, highlights how dependent actions can be mistakenly thought of as independent when we multiply their probabilities to find the sequence outcome: taking the first probability and multiplying it with the adjusted second one. Always check whether the first action changes the conditions for the second when dealing with probabilities.