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Chapter 1: Problem 16

A coin is tossed two independent times, each resulting in a tail \((\mathrm{T})\) or a head (H). The sample space consists of four ordered pairs: TT, TH, HT, HH. Making certain assumptions, compute the probability of each of these ordered pairs. What is the probability of at least one head?

Short Answer

Expert verified
The probability of each ordered pair (TT, TH, HT, HH) is 0.25. The probability of getting at least one head is 0.75.

Step by step solution

01

Compute Probability of Each Ordered Pair

To compute the probability of an ordered pair is as simple as multiplying the probabilities of the individual outcomes. Since these are fair coin tosses, the probability of getting a head (H) or a tail (T) in each toss is 0.5. For example, to compute the probability of TT (both tosses yield tails), we do: \[ \text{P(TT)} = P(T) * P(T) = 0.5 * 0.5 = 0.25 \] Similarly, we can compute the probabilities for other three outcomes (TH, HT, HH). They are all 0.25.
02

Determine the Probability of at Least One Head

To determine the probability of getting at least one head in two coin tosses, we can sum up probabilities of the outcomes that include at least one head. That means, we account for TH, HT, HH outcomes. So, the probability of getting at least one head is: \[ \text{P(at least one H)} = P(TH) + P(HT) + P(HH) = 0.25 + 0.25 + 0.25 = 0.75 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Space
When dealing with probability theory, it is essential to understand the concept of sample space. Sample space is the set of all possible outcomes in a particular experiment. In the context of our exercise, where a coin is tossed two times, each toss can result in one of two outcomes: heads (H) or tails (T).

With two tosses, we must consider all combinations of these outcomes. Each outcome from the experiment can be represented as an ordered pair. Therefore, the sample space consists of four outcomes:
  • TT
  • TH
  • HT
  • HH
TT represents both tosses resulting in a tail, TH represents the first toss being tails and the second heads, etc. Understanding this concept helps in evaluating probabilities since it's the foundation for all calculations.
Independent Events
In probability theory, independent events are those whose occurrences do not affect each other. In simpler terms, the result of one event has no bearing on the result of another.

For example, when tossing a coin, whether you get heads or tails does not influence the result of the next toss. Each coin toss is an independent event, which is crucial for calculating probabilities accurately.

To find the probability of two independent events happening together, you multiply the probabilities of each event. If event A is getting a tail, and event B is getting a tail again on the second toss, both with a probability of 0.5, the probability of TT is \( P(T) \times P(T) = 0.5 \times 0.5 = 0.25 \).
This principle holds true for all other pairs in our coin toss exercise.
Fair Coin Toss
A fair coin toss is a fundamental concept in probability, representing an experiment where a coin is tossed in such a manner where it has an equal chance of landing on heads or tails. This is often used in probability studies due to the simplicity and clarity it provides in understanding randomness.

When dealing with a fair coin, each side (heads or tails) has a probability of 0.5 on any given toss. This equality ensures that each outcome is equally likely, making fair coins essential tools in teaching probability theory concepts.

In our exercise, this fair nature directly influences the calculations. For example, both heads and tails have the same chance of occurring at each toss.
  • This results in all four sample space outcomes (TT, TH, HT, HH) having equal probabilities of 0.25.
  • This uniform probability distribution aids in easily determining composite outcome probabilities, such as the probability of getting at least one head.
Understanding the fair coin toss concept is crucial for accurate analysis and to ensure unbiased results in probability experiments.

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Most popular questions from this chapter

A drawer contains eight different pairs of socks. If six socks are taken at random and without replacement, compute the probability that there is at least one matching pair among these six socks. Hint: Compute the probability that there is not a matching pair.

In a certain factory, machines I, II, and III are all producing springs of the same length. Machines I, II, and III produce \(1 \%, 4 \%\), and \(2 \%\) defective springs, respectively. Of the total production of springs in the factory, Machine I produces \(30 \%\), Machine II produces \(25 \%\), and Machine III produces \(45 \%\). (a) If one spring is selected at random from the total springs produced in a given day, determine the probability that it is defective. (b) Given that the selected spring is defective, find the conditional probability that it was produced by Machine II.

From a bowl containing five red, three white, and seven blue chips, select four at random and without replacement. Compute the conditional probability of one red, zero white, and three blue chips, given that there are at least three blue chips in this sample of four chips.

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 \(P\left(C_{1} \cup C_{2}\right)\).

For each of the following cdfs \(F(x)\), find the pdf \(f(x)[\mathrm{pmf}\) in part \((\mathbf{d})]\), the first quartile, and the \(0.60\) quantile. Also, sketch the graphs of \(f(x)\) and \(F(x)\). May use \(\mathrm{R}\) to obtain the graphs. For Part(a) the code is provided. (a) \(F(x)=\frac{1}{2}+\frac{1}{\pi} \tan ^{-1}(x),-\infty
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