/*! 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 18 If you flip a coin ten times, yo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If you flip a coin ten times, you expect on average to get five heads and five tails. a. The pattern HHHHHHHHHH violates this expectation dramatically. What is the probability of obtaining this pattern? b. The pattern HTHTHTHTHT matches this expectation exactly. What is the probability of obtaining this pattern? c. What is the probability of obtaining the pattern HTTTHHTTHT? d. What is the probability of obtaining a pattern with one tail and nine heads?

Short Answer

Expert verified
a. \( \frac{1}{1024} \) b. \( \frac{1}{1024} \) c. \( \frac{1}{1024} \) d. \( \frac{10}{1024} \)

Step by step solution

01

Define Total Flips

Each coin flip has two outcomes: heads (H) or tails (T). When flipping a coin 10 times, there are a total of \(2^{10}\) possible outcomes, because each flip has 2 outcomes.
02

Calculate Probability of HHHHHHHHHH

The pattern HHHHHHHHHH consists only of heads for each of the 10 flips. There is exactly 1 way to achieve this pattern out of the total \(2^{10}\) possible outcomes.Thus, the probability \( P(\text{HHHHHHHHHH}) = \frac{1}{2^{10}} \).
03

Calculate Probability of HTHTHTHTHT

The pattern HTHTHTHTHT is one specific sequence of flips, alternating heads and tails. There is exactly 1 way to achieve this pattern.Thus, the probability \( P(\text{HTHTHTHTHT}) = \frac{1}{2^{10}} \).
04

Calculate Probability of HTTTHHTTHT

The pattern HTTTHHTTHT is one specific sequence among the 1024 possible outcomes.Thus, the probability \( P(\text{HTTTHHTTHT}) = \frac{1}{2^{10}} \).
05

Calculate Probability of 1 Tail and 9 Heads

To find the probability of obtaining 1 tail and 9 heads, we first determine the number of ways to arrange 9 heads and 1 tail in 10 flips. This is the number of combinations of 10 items taken 1 at a time, which is given by \( \binom{10}{1} = 10 \).Thus, the probability of having 9 heads and 1 tail is \( P(\text{1 T and 9 H}) = \frac{10}{2^{10}} \).

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.

Coin Flipping
When it comes to probability, coin flipping is one of the simplest examples. Each flip of a coin has two possible outcomes: heads (H) or tails (T). These are often seen as perfectly random events. Each outcome is equally likely, meaning the probability of getting a heads is the same as getting a tails, which is 0.5, or 50%. If you flip a coin once, there's a 50% chance it will land on heads and a 50% chance it will land on tails. Now imagine flipping the coin multiple times. Things start to get more interesting. If you flip a coin 10 times, for example, you might expect around five heads and five tails. But randomness can sometimes give outcomes that deviate from this expectation, just as in the original exercise. Some patterns, like ten heads in a row, can occur, but they do so with a very low probability because there is only one such sequence among the 1024 possible results.
Combinatorics
Combinatorics is a branch of mathematics that deals with counting combinations and permutations. In the context of coin flipping, it helps us determine the number of possible sequences or outcomes. If you're interested in specific patterns from flips, combinatorics is your friend. For example, consider the pattern of obtaining exactly 1 tail out of 10 flips:
  • The total number of ways to arrange 1 tail and 9 heads in a sequence of 10 flips can be determined using the combination formula, denoted as \( \binom{n}{k} \).
  • Here, \( \binom{10}{1} \) reflects we choose 1 position for a tail out of 10 possible positions.
  • This concept helps in visualizing how complex sequences can come together through basic counting techniques.
Binomial Probability
Binomial probability plays a key role when dealing with fixed numbers of sequential trials, like coin flips. In a scenario where each flip independently results in heads or tails, we consider binomial trials. The probability of obtaining exactly k successes (in this case, heads) in n trials (flips) is calculated using the binomial probability formula: \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Where:
  • \( n \) is the number of trials (flips)
  • \( k \) is the number of successful trials (desired outcome, e.g., heads)
  • \( p \) is the probability of success on a single trial
The formula gives a systematic way to calculate the likelihood of getting a certain number of heads in a given set of flips, allowing for calculation beyond mere gut feeling.
Conditional Probability
Conditional probability is a critical concept that helps us understand probabilities in the context of known conditions. In this term, it’s defined as the probability of an event occurring given that another event has already occurred. This idea isn't directly addressed in the exercise, as it more concerns standalone probabilities. Nonetheless, understanding the basis provides depth to probability theory. For example, if we knew that the first 5 flips resulted in heads, the probability of the remaining flips produces a more nuanced outcome. Essentially, conditional probability provides a tool to refine probability estimates based on updated information. It's very handy when evaluating probabilities as new data or outcomes are revealed step-by-step. It allows for a more granular understanding of outcomes by not treating them as isolated events but linked to preceding circumstances.

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

(E) Accessible regions of phase space Suppose that \(N\) non-interacting particles, each of mass \(m\), move freely in a one-dimensional box (i.e. an infinite square well). Denote the position coordinates by \(x_{1}, x_{2}, \ldots, x_{N}\) and the momentum coordinates by \(p_{1}, p_{2}, \ldots, p_{N}\). The box restricts all the positions to fall between \(x_{i}=0\) and \(x_{i}=L .\) The energy of the system lies between \(E\) and \(E+\Delta E\). a. If only one particle is present, draw the system's phase space and shade the regions of phase space that are accessible. b. If two particles are present then phase space is four dimensional, which makes it difficult to draw. Draw separately the part of phase space involving positions and the part involving momenta. Shade the accessible regions of phase space. c. Suppose two particles are present, and consider the slice of phase space for which \(x_{1}=(2 / 3) L\) and \(p_{2}\) equals some constant called \(\bar{p}_{2}\). Draw a (carefully labeled) sketch of this slice with the accessible regions shaded. d. Describe the accessible regions of phase space if \(N\) particles are present.

(Q) For discussion: Mechanical parameters, dynamical variables, and modeling List the mechanical parameters and dynamical variables of these systems: a. Hydrogen molecules enclosed in a sphere. b. Water molecules in a box. c. A mixture of hydrogen molecules and helium atoms in a box. To what extent are you making models as you generate descriptions? To what extent are you making assumptions? (For example, by using non-relativistic classical mechanics.) \({ }^{2}\) For the earth-moon model, all the points visited will also have the same total momentum and angular momentum, but this is not the case for the helium-in-a-smooth-box model.

(E) The logarithm Suppose that a differentiable function satisfies $$ f(x y)=f(x)+f(y) $$ for all positive \(\mathrm{x}\) and \(\mathrm{y}\). Show that $$ f(x)=k \ln (x) $$ [Clues: 1) Take derivative with respect to \(x\), then set \(x=1.2\) ) Set \(y=1\) in Equation 2.4.4.]

(Q,E) Lost in space A collection of \(\mathrm{N}\) asteroids floats in space far from other gravitating bodies. Model each asteroid as a hard sphere of radius \(\mathrm{R}\) and mass \(\mathrm{m}\). What quantities are required for a microscopic description of this system? For a macroscopic description?

(Q) Three-body interactions: macroscopic Even though three-body interactions do exist, they can usually be ignored in a macroscopic description. (Just as the wall-atom interaction does exist, but it can usually be ignored in a macroscopic description.) Why?

See all solutions

Recommended explanations on Physics 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.