Question

A coin is to be tossed as many times as necessary to turn up one head. Thus the elements \(c\) of the sample space \(\mathcal{C}\) are \(H, T H, T T H, T T T H\), and so forth. Let the probability set function \(P\) assign to these elements the respec- tive probabilities \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\), and so forth. Show that \(P(\mathcal{C})=1 .\) Let \(C_{1}=\\{c\) : \(c\) is \(H, T H, T T H, T T T H\), or \(T T T T H\\} .\) Compute \(P\left(C_{1}\right) .\) Next, suppose that \(C_{2}=\) \(\left\\{c: c\right.\) is TTTTH or TTTTTH\\}. Compute \(P\left(C_{2}\right), P\left(C_{1} \cap C_{2}\right)\), and \(P\left(C_{1} \cup C_{2}\right)\).

Short answer

P(C) = 1, P(C1) = 31/32, P(C2) = 3/64, P(C1 ∩ C2) = 1/32, P(C1 ∪ C2) = 31/32

Step-by-step solution

Compute P(C)

Calculate the total probability of the sample space \(\mathcal{C}\). This is simply the sum of the probabilities of each event occurring. These events are mathematically represented as a geometric progression series since each probability value is half of the previous one. The sum of an infinite geometric progression series with a first term \(a\) and common ratio \(r\) (where \(|r|<1\)) is \(S=a/(1-r)\). Here, \(a = 1/2\) and \(r = 1/2\), so \(S = (1/2) / (1 - 1/2) = 1\). Therefore, the total probability \(P(\mathcal{C}) = 1\).

Compute P(C1)

The set \(C1\) includes the outcomes {H, TH, TTH, TTTH, TTTTH}. To compute the probability of \(C1\), you add the probabilities of each of these outcomes, which are respectively: \(1/2, 1/4, 1/8, 1/16, 1/32\). Summing these up gives the result \(P(C1) = 31/32\).

Compute P(C2)

The set \(C2\) includes the outcomes {TTTTH, TTTTTH}. Their respective probabilities are \(1/32\) and \(1/64\), so \(P(C2) = 1/32 + 1/64 = 3/64\).

Compute P(C1 ∩ C2) and P(C1 ∪ C2)

\(P(C1 ∩ C2)\) is the probability of the event \(C1\) and \(C2\) occurring together. The intersection of \(C1\) and \(C2\) is {TTTTH}, whose probability is \(1/32\), so \(P(C1 ∩ C2) = 1/32\). The union \(P(C1 ∪ C2)\) involves calculating the probability of either \(C1\) or \(C2\) or both occurring. This is a case where the principle inclusion-exclusion is used: \(P(C1 ∪ C2) = P(C1) + P(C2) - P(C1 ∩ C2) = 31/32 + 3/64 - 1/32 = 31/32\).

Key concepts

Geometric Progression Series

Understanding geometric progressions is critical when analyzing probabilities in scenarios like coin tossing, where outcomes follow a pattern of being a certain fraction of the probability of the previous outcome. A geometric progression series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio.

In probability calculations like the coin toss, we often sum an infinite geometric series. The formula for the sum of an infinite geometric progression where the common ratio's absolute value is less than one is \[ S = \frac{a}{1 - r} \. \] Here, 'a' stands for the first term in the series, and 'r' is the common ratio. The essence of the geometric progression in probability is to assert that because each subsequent event is less likely to happen by a factor of 'r', the total sum converges to a finite number, ensuring that the total probability of all possible outcomes is 1.

Total Probability

The concept of total probability is a vital foundation in understanding how to approach problems involving probability spaces or sample sets. It suggests that the sum of probabilities for all possible exclusive outcomes of a random experiment must equal 1. This is known as the normalization condition of probability.

In the context of the textbook coin toss exercise, each coin toss is an independent event, and the series of tosses leading to the first head is exhaustive and mutually exclusive. This means the probability of the entire set, denoted by \( P(\mathcal{C}) \), must add up to 1. Using the geometric series property, where each term is a fraction of the preceding probability, ensures compliance with the rule of total probability, confirming that the model of the coin toss example accurately represents a valid probability space.

Inclusion-Exclusion Principle

When dealing with sets in probability, sometimes an event can be a part of multiple sets. The inclusion-exclusion principle enables us to find the probability of the union of events by appropriately adding and subtracting the probabilities of individual events and their intersections.

The formula for two sets A and B, for example, goes as follows: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \.\] It ensures that we don't overcount probabilities when two events have common outcomes. In our coin toss exercise, this principle is used to find \( P(C1 \cup C2) \), where it is apparent that while adding the probabilities of \( C1 \) and \( C2 \), we should subtract the probability of \( C1 \cap C2 \) since that event is included in the probabilities of both \( C1 \) and \( C2 \.\)

Sample Space

In probability theory, the sample space is the set of all possible outcomes of a random experiment. It is denoted by a capital letter, such as \( C \) in our coin toss example, and each outcome within the sample space is an element of that space. Understanding the structure of the sample space is essential for defining events and calculating their probabilities.

In the coin toss scenario, the sample space includes infinite elements like H, TH, TTH, and so on. This exercise required that you comprehend the idea of a theoretically infinite sample space while also realizing that it results in finite probabilities, a sometimes counterintuitive concept that lies at the heart of probability theory. By grasping the sample space and how it relates to events and their probabilities, students can better solve and understand similar probability problems.