/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Suppose we flip a fair coin four... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose we flip a fair coin four times. We are interested in counting the number of times the coin turns up heads. (a) Define a sample space \(\Omega\) and a probability density \(p\) on \(\Omega\). (b) Define a random variable \(X\) on \(\Omega\) to count the number of heads. (c) Describe the event \((X=3)\) as a subset of \(\Omega\). (d) Set up a sample space \(\Omega_{x}\) and a probability distribution \(p_{x}\) based on \(X\). You may express your answer in terms of the binomial distribution. (c) Which number (or numbers) of heads is most likely to occur?

Short Answer

Expert verified
2 heads is the most likely number.

Step by step solution

01

Define the Sample Space

The sample space \(\Omega\) for flipping a fair coin four times consists of all possible outcomes of the sequences of heads (H) and tails (T). Since each flip has 2 possible outcomes, \(\Omega\) will have \(2^4 = 16\) elements. These are: \(\{\text{HHHH}, \text{HHHT}, \text{HHTH}, \text{HHTT}, \text{HTHH}, \text{HTHT}, \text{HTTH}, \text{HTTT}, \text{THHH}, \text{THHT}, \text{THTH}, \text{THTT}, \text{TTHH}, \text{TTHT}, \text{TTTH}, \text{TTTT}\}\). Each outcome is equally likely, so the probability of each is \(p(\omega) = \frac{1}{16}\) for all \(\omega \in \Omega\).
02

Define the Random Variable X

Define the random variable \(X\) on \(\Omega\) that counts the number of heads. For any \(\omega \in \Omega\), \(X(\omega)\) is the number of 'H' in the outcome. For example, if \(\omega = \text{HHHT}\), then \(X(\omega) = 3\).
03

Describe the Event (X=3)

The event \((X=3)\) represents the subset of \(\Omega\) where exactly three heads occur. This subset is \(\{\text{HHHT}, \text{HHTH}, \text{HTHH}, \text{THHH}\}\).
04

Set up a Sample Space and Probability Distribution based on X

The sample space \(\Omega_{x}\) for the random variable \(X\) consists of the possible values \(X\) can take, which are \{0, 1, 2, 3, 4\}. The probability distribution \(p_{x}\) based on \(X\) follows a binomial distribution with parameters \(n=4\) and \(p=0.5\), given by \(p_{x}(k) = \binom{4}{k} (0.5)^k (0.5)^{4-k}\).
05

Determine the Most Likely Number of Heads

The number of heads that is most likely to occur can be found by determining the mode of the binomial distribution from Step 4. The mode (most probable number of heads) of a binomial distribution when \(n=4\) and \(p=0.5\) is closest to \(np = 2\). Therefore, 2 heads are the most likely outcome.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Space
In probability theory, a sample space is a collection of all possible outcomes of an experiment. Let's consider our scenario of flipping a fair coin four times. Each flip has two possible outcomes: heads (H) or tails (T). Therefore, if we flip the coin four times, there are a total of \(2^4 = 16\) possible sequences of heads and tails. These make up our sample space \(\Omega\). Each sequence is an element of \(\Omega\), for example, "HHHT" or "TTHH".

Since the coin is fair, each of these outcomes is equally probable. This means each sequence has a probability of \(\frac{1}{16}\). Hence, the sample space is fundamental in understanding the complete set of scenarios that can occur in our experiment, helping us calculate probabilities effectively.
Random Variable
A random variable is a function that assigns a numerical value to each element in a sample space. It helps us quantify the outcomes of a probabilistic event. In our scenario, we define a random variable \(X\) that counts the number of times the coin shows heads when flipped four times.

For any specific outcome \(\omega\) in \(\Omega\), \(X(\omega)\) will denote the number of heads in that outcome. For example, if the outcome is "HHHT", then \(X(\omega) = 3\) since there are 3 heads. The random variable \(X\) provides a way to connect the outcomes with numeric values, making it easier to calculate probabilities for specific counts of heads.
Binomial Distribution
The binomial distribution is a type of probability distribution that summarizes the likelihood of achieving a specific number of successes in a fixed number of independent trials, each with the same probability of success. In our coin flipping experiment, the success is defined as flipping a head.

The parameters of the binomial distribution here are \(n = 4\) (the number of coin flips) and \(p = 0.5\) (the probability of flipping a head). The probability of getting exactly \(k\) heads in four flips follows the formula \[p_{x}(k) = \binom{4}{k} (0.5)^k (0.5)^{4-k}\].
Each \(k\) value reflects a different condition, for instance, \(\{0, 1, 2, 3, 4\}\), showing the probabilities for 0 to 4 heads respectively. This framework of the binomial distribution facilitates predicting the probability of various outcomes.
Probability Distribution
A probability distribution describes how the probabilities are distributed over possible outcomes of a random variable. For our random variable \(X\) that counts the number of heads obtained, the probability distribution gives us probabilities of obtaining \(0, 1, 2, 3,\) and \(4\) heads.

Given our binomial distribution setup with \(n = 4\) and \(p = 0.5\), we can calculate the probabilities for each count of heads. This gives us a clear understanding of what to expect after all trials. For instance, the distribution shows that achieving 2 heads is the most probable outcome, known as the mode. Probability distributions help in visualizing and predicting outcomes in experiments involving chance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that \(\Omega\) is a sample space with a probability density function \(p\), and suppose that \(A \subseteq \Omega\). Let \(P(A)\) denote the probability of \(A\). Assume that \(P(A)>0 .\) Define a function \(p_{1}\) on \(A\) as follows: For \(\omega \in A, p_{1}(\omega)=p(\omega) / P(A)\). (a) Show that if \(\omega_{1}, \omega_{2} \in A\) and \(p\left(\omega_{1}\right), p\left(\omega_{2}\right) \neq 0,\) then \(\frac{p\left(\omega_{1}\right)}{p\left(\omega_{2}\right)}=\frac{p_{1}\left(\omega_{1}\right)}{p_{1}\left(\omega_{2}\right)}\) (b) Show that if \(B\) and \(C\) are nonempty subsets of \(A\) with elements that have positive probabilities, then \(\frac{P(B)}{P(C)}=\frac{P_{1}(B)}{P_{1}(C)}\) (c) Show that \(p_{1}\) is a probability density function on \(\Omega_{1}=A\).

A computer salesperson makes either one or two sales contacts each day between 1 and 2 PM. If only one contact is made, the probability is 0.2 that a sale will result and 0.8 that no sale will result. If two contacts are made, the two customers make their decisions independently of each other, each purchasing with probability 0.2 and not purchasing with probability 0.8 . What is the probability that the salesperson has made two sales this hour?

Suppose that \(E_{1}, E_{2}, \ldots, E_{k}\) are events in the same sample space and that some pair \(E_{i}, E_{j}\) of these events are disjoint. (a) If all the events have positive probability, can the set \(\left\\{E_{1}, E_{2}, \ldots, E_{k}\right\\}\) be an independent set of events? Explain your answer. (b) If one or more of the events has 0 probability, can the set \(\left\\{E_{1}, E_{2}, \ldots, E_{k}\right\\}\) be an independent set of events?

Let a random variable \(X\) have probability density function $$\begin{array}{l|c|c|c|c}x & 1 & 2 & 6 & 8 \\\\\hline p(X=x) & 0.4 & 0.1 & 0.3 & 0.2\end{array}$$ Compute the variance and standard deviation of \(X\) with \(\mu=4\).

A fair die is rolled, and a fair coin is tossed. The sample space is taken to be \(\Omega=\) \(\Omega_{1} \times \Omega_{2}\) where \(\Omega_{1}\) is the six- clement sample space for the die and \(\Omega_{2}\) is the twoelement sample space for the coin. Let \(A \subseteq \Omega_{1}\) be the event "a 5 is rolled." Let \(B \subseteq \Omega_{2}\) be the event "heads." Let \(C \subseteq \Omega\) be the event "at most two spots on the top face of the die (with heads or tails on the coin) or at least five spots on the top face of the die together with heads on the coin." Let \(D\) be the event "at least a 5 on the die (with heads or tails on the coin)." Which of the following sets of events are independent sets? Explain your answer. (a) \(\\{A, B\\}\) (b) \(\\{A, B, C\\}\) (c) \(\\{B, C]\) (d) \(\\{B, C, D\\}\)

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.