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 \(H, T H, T T H\), TTTH, and so forth. Let the probability set function \(P\) assign to these elements the respective 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 \(H, T H, T T H\), TTTH, or TTTTH \(\\}\). Compute \(P\left(C_{1}\right)\). Next, suppose that \(C_{2}=\) \(\\{c: c\) 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

The probabilities are $P\left(C_{1}\right)=\frac{31}{32}$, $P\left(C_{2}\right)=\frac{3}{32}$, $P\left(C_{1} \cap C_{2}\right)=\frac{3}{32}$ and $P\left(C_{1} \cup C_{2}\right)=\frac{31}{32}$.

Step-by-step solution

Identify the sample space and its probability

The sample space consists of H, TH, TTH, TTTH, etc. and the probability assigned to these outcomes follows a geometric progression with a common ratio of \(\frac{1}{2}\). Therefore, the sum of this infinite geometric series equals \(1\). This validates that $P(\mathcal{C})=1$.

Compute the probability for event C1

Event space \(C1\) includes H, TH, TTH, TTTH, and TTTTH. We have to sum the probabilities of these outcomes: \\[P(C1) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \] \This gives us $P\left(C_{1}\right)=\frac{31}{32}$.

Compute the probability for event C2

Event space \(C2\) includes TTTTH and TTTTTH. We can sum the probabilities of these outcomes: \\[P(C2) = \frac{1}{16} + \frac{1}{32} \] \which sums up to $P\left(C_{2}\right)=\frac{3}{32}$.

Compute the probability of the intersection of events C1 and C2

The intersections of the events C1 and C2 will be elements that are common in both of the event spaces. The common outcomes are TTTTH and TTTTTH. Therefore, \\[P(C1 ∩ C2) = \frac{1}{16} + \frac{1}{32} \] \Which gives us $P\left(C_{1} \cap C_{2}\right)=\frac{3}{32}$.

Compute the probability of the union of events C1 and C2

The union of the events includes all the outcomes from both event spaces C1 and C2. \Notice that $C_{1} \cup C_{2}=C_{1}$ because all elements of $C_{2}$ are also elements of $C_{1}$. Therefore, $P\left(C_{1} \cup C_{2}\right)=P\left(C_{1}\right)=\frac{31}{32}$.

Key concepts

Infinite Geometric Series

When a coin toss continues indefinitely until a head comes up, each sequence of tails followed by a head forms part of an infinite geometric series. A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a constant called the ratio. In this case, the ratio is \frac{1}{2}, as the chance of getting a tail and then a head is half that of just getting a head, and so on for longer sequences of tails.

Summing the infinite geometric series allows us to determine the total probability for the complete sample space, which logically should be equal to 1 since one of the outcomes must occur. The formula for the sum S of an infinite geometric series with the first term a and the common ratio r (where \r| < 1) is given by \[S = \frac{a}{1 - r}\]
In our example, a = \frac{1}{2} (the probability of getting a head on the first toss) and r = \frac{1}{2}. Plugging these values into the formula gives \[S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1\], confirming that the complete set of outcomes covers all possible scenarios (hence the sum of probabilities is 1).

Sample Space in Probability

The sample space in a probability scenario is the set of all possible outcomes of a random experiment. In the context of tossing a coin until a head appears, the sample space is comprised of sequences like H, TH, TTH, TTTH, etc., where 'T' represents tails, and 'H' represents heads. It's essential to define the sample space correctly to ensure that we account for all possible outcomes when computing probabilities.

Each outcome in the sample space is assigned a probability, and the sum of these probabilities must equal 1, the certainty of any outcome happening. In the given exercise, we're shown that this sum equals 1 by using the properties of an infinite geometric series, solidifying our understanding that we're covering the entirety of the sample space with our event outcomes. Ultimately, defining the sample space correctly is pivotal as it forms the basis for all subsequent probability calculations.

Probability Set Function

A probability set function assigns a probability to each event in a given sample space, and it adheres to specific axioms such as non-negativity, normalization, and additivity. In our exercise, the function is defined to give probabilities of \(\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\), and so on, for the respective outcomes, squarely fitting the description of a probability set function.

The function's normalization property ensures that the sum of probabilities of all possible outcomes is 1, reflecting the absolute certainty that one of the outcomes in the sample space will occur. The additivity axiom is illustrated when we calculate probabilities for events like \(C1\) and \(C2\), as well as their intersections and unions. The probability of an event is the sum of the probabilities of the individual outcomes that constitute the event. For instance, the probability of the event \(C1\) is computed by adding the probabilities of getting a head in the first, second, third, fourth, or fifth toss, leading to a probability sum of \(P(C1) = \frac{31}{32}\).

When events are mutually exclusive, the probability of their union is simply the sum of their individual probabilities. However, if they are not, such as the cases of \(C1\) and \(C2\) in our exercise, we subtract the probability of their intersection from the sum of their individual probabilities to avoid double-counting. This attention to detail ensures that probabilities are assigned consistently and accurately, based on the principles of probability theory.