Problem 44 - Algebra and Trigonometry

Question

A coin is tossed twice. Let \(E\) and \(F\) be the following events: $$\begin{array}{l}{E : \text { The first toss shows heads }} \\ {F : \text { The second toss shows heads }}\end{array}$$ (a) Are the events \(E\) and \(F\) independent? (b) Find the probability of showing heads on both tosses.

Step-by-Step Solution

Answer

(a) Yes, events \(E\) and \(F\) are independent. (b) Probability of heads on both tosses is \(\frac{1}{4}\).

Step 1: Identify Given Events

The events are defined based on the outcomes of tossing a coin twice. Event \(E\) occurs if the first toss results in a heads, and event \(F\) occurs if the second toss results in a heads.

Step 2: Determine Sample Space

When a coin is tossed twice, the sample space of possible outcomes is \( \{HH, HT, TH, TT\} \), where \(H\) denotes heads and \(T\) denotes tails.

Step 3: Calculate Probability of Each Event

Determine the probability of each event using the sample space. For event \(E\), the outcomes are \( \{HH, HT\} \), so \(P(E) = \frac{2}{4} = \frac{1}{2}\). For event \(F\), the outcomes are \( \{HH, TH\} \), so \(P(F) = \frac{2}{4} = \frac{1}{2}\).

Step 4: Calculate Joint Probability

Determine the probability of both events \(E\) and \(F\) occurring together, which occurs only when the outcome is \(HH\). Thus, \(P(E \cap F) = \frac{1}{4}\).

Step 5: Check Independence of Events

The events \(E\) and \(F\) are independent if \(P(E \cap F) = P(E) \times P(F)\). Here, \(P(E) \times P(F) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\), which matches \(P(E \cap F) = \frac{1}{4}\). So, \(E\) and \(F\) are independent.

Step 6: Find Probability of Heads on Both Tosses

The probability of getting a head on both tosses is simply \(P(E \cap F)\), which was calculated as \(\frac{1}{4}\) in the previous step.

Key Concepts

Sample Space

In probability theory, the concept of a sample space is fundamental. It represents the set of all possible outcomes of a random experiment. When we toss a coin twice, the sample space includes every combination of what might happen: heads or tails on each toss. For our specific exercise, the sample space is denoted by the set \( \{HH, HT, TH, TT\} \). Here, \(H\) signifies a head, while \(T\) stands for a tail.

Understanding the sample space is essential as it lays the groundwork for calculating probabilities. Each possible outcome in the sample space is equally likely when a fair coin is used. This means that each of the four outcomes—"HH," "HT," "TH," and "TT"—is expected to occur with a probability of \( \frac{1}{4} \).

Knowing the sample space helps us identify specific events and their probabilities. For instance, if we want to find the probability of obtaining at least one head in two tosses, we would look for the occurrences "HH," "HT," or "TH" in our sample space.

Independent Events

An important concept in probability theory is that of independent events. Two events are independent if the occurrence of one event does not affect the probability of the other event occurring. For our problem, these events are the result of the first and second coin toss.

• Event \(E\): Getting heads on the first toss.
• Event \(F\): Getting heads on the second toss.

To check whether events \(E\) and \(F\) are independent, we can use the formula \(P(E \cap F) = P(E) \times P(F)\). We find that the joint probability \(P(E \cap F)\) is \(\frac{1}{4}\), which matches the product \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\).

This result confirms that the events are indeed independent. This means that obtaining a head on the first toss has no impact on whether a head will appear on the second toss.

Joint Probability

Joint probability is the probability that two events will occur simultaneously. In our coin toss scenario, it involves both the first toss and the second toss resulting in heads.

• Event \(E \cap F\): Both the first and second toss landing on heads ("HH").

To compute \(P(E \cap F)\), we look at the sample space and see that one of the four events corresponds to "HH". Therefore, \(P(E \cap F) = \frac{1}{4}\).

The joint probability is useful to determine the co-occurrence of events and plays a crucial role in understanding their interdependence. In this exercise, calculating \(P(E \cap F)\) helped establish the independence of events \(E\) and \(F\).

Coin Toss

A coin toss is a simple yet powerful example to illustrate fundamental principles of probability. Each toss of a coin has only two possible outcomes: heads (H) or tails (T). This makes it a classic model for understanding randomness and chance.

• With one toss, there is a probability of \(\frac{1}{2}\) for either heads or tails.
• When tossing twice, the task becomes a bit more complex due to the increased number of possible outcomes.

By calculating probabilities for events based on multiple coin tosses, we can explore concepts like sample space, independence, and joint probability. A fair coin guarantees that outcomes are unbiased, making it a cornerstone for demonstrating key ideas in probability theory.

This exercise exemplifies how a simple act like tossing a coin can open up a deeper understanding of the principles governing event probabilities.