概率论基础教程：第8版

probability

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A first course in probability

A first course in probability(概率论基础教程)

Ross的A first course in probability(概率论基础教程)第七版英文版，djvu格式的，需要自己转成pdf，看起来比较费劲

A first course in probability 8th Edith

英文版概率论第八版，经典教材，可与已上传的中文版结合使用，PDF，解压即可

An urn initially contains 5 white and 7 black balls. Each time a ball is selected, its color is noted and it is replaced in the urn along with 2 other balls of the same color. Compute the probability that a. the first 2 balls selected are black and the next 2 are white; b. of the first 4 balls selected, exactly 2 are black.

a. 首先抽出2个黑球和2个白球的概率为：7/12 * 6/13 * 5/15 * 4/17 = 0.096。 b. 首先抽出2个黑球和2个白球的概率为：7/12 * 6/13 * 5/15 * 4/17 = 0.096。

帮我补充代码：class LanguageModel(LanguageModel): def get_unigram_logp(self, unigram): """Computes the log-probability of unigram under this LanguageModel. Args: unigram (str): Unigram for which to compute the log-probability. Returns: log_p (float): Log-probability of unigram under this LanguageModel. """ ### Begin your code ### End your code def get_bigram_logp(self, w_1, w_2): """Computes the log-probability of unigram under this LanguageModel. Note: Use self.lambda_ for the unigram-bigram interpolation factor. Args: w_1 (str): First word in bigram. w_2 (str): Second word in bigram. Returns: log_p (float): Log-probability of bigram under this LanguageModel. """ ### Begin your code ### End your code def get_query_logp(self, query): """Computes the log-probability of query under this LanguageModel. Args: query (str): Whitespace-delimited sequence of terms in the query. Returns: log_p (float): Log-probability assigned to the query under this LanguageModel. """ ### Begin your code ### End your code

unigram (str): Unigram for which to compute the log-probability. Returns: log_p (float): Log-probability of unigram under this LanguageModel. """ count_w1 = self.unigram_counts[unigram] total_...

AttributeError: module 'tensorflow_probability.python.bijectors.bijector' has no attribute 'AutoCompositeTensorBijector'

在较旧的版本中，'tensorflow_probability.python.bijectors.bijector'模块中没有'AutoCompositeTensorBijector'属性。建议您更新tensorflow_probability到最新版本，可以使用以下命令进行更新： pip install --...

Excess probability

Excess probability is the probability of an event occurring that is above the expected or baseline probability. It is the difference between the actual probability and the expected probability of an ...

Error in nomogram(data = df, fun = function(x) 0.5, varname = c("original.firstorder.RobustMeanAbsoluteDeviation", : argument "fit" is missing, with no default

`original:firstorder:RobustMeanAbsoluteDeviation` = c(0, 0.5, 1), `log-sigma-4-0-mm-3D:firstorder:InterquartileRange` = c(0, 0.5, 1), `log-sigma-5-0-mm-3D:glszm:SmallAreaEmphasis` = c(0, 0.5, 1), ...

import numpy as np import matplotlib.pyplot as plt import math def count(lis): lis = np.array(lis) key = np.unique(lis) x = [] y = [] for k in key: mask = (lis == k) list_new = lis[mask] v = list_new.size x.append(k) y.append(v) return x, y mu = [14, 23, 22] sigma = [2, 3, 4] tips = ['design', 'build', 'test'] figureIndex = 0 fig = plt.figure(figureIndex, figsize=(10, 8)) color = ['r', 'g', 'b'] ax = fig.add_subplot(111) for i in range(3): x = np.linspace(mu[i] - 3sigma[i], mu[i] + 3sigma[i], 100) y_sig = np.exp(-(x - mu[i])**2/(2*sigma[i]**2))/(math.sqrt(2*math.pi)) ax.plot = (x, y_sig, color[i] + '-') ax.legend(loc='best', frameon=False) ax.set_xlabel('# of days') ax.set_ylabel('probability') plt.show() plt.grid(True) size = 100000 samples = [np.random.normal(mu[i], sigma[i], size) for i in range(3)] data = np.zeros(len(samples[1])) for i in range(len(samples[1])): for j in range(3): data[i] += samples[j][i] data[i] = int(data[i]) a, b = count(data) pdf = [x/size for x in b] cdf = np.zeros(len(a)) for i in range(len(a)): if i > 0: cdf[i] += cdf[i - 1] cdf = cdf/size figureIndex += 1 fig = plt.figure(figureIndex, figsize=(10, 8)) ax = fig.add_subplot(211) ax.bar(a, height=pdf, color='blue', edgecolor='white', label='MC PDF') ax.plot(a, pdf) ax.legend(loc='best', frameon=False) ax.set_xlabel('# of days for project') ax.set_ylabel('probability') ax.set_title('Monte Carlo Simulation') ax = fig.add_subplot(212) ax.plot(a, cdf) ax.legend(loc='best', frameon=False) ax.set_xlabel('# of days for project') ax.set_ylabel('probability') ax.grid(True) plt.show()修改一下代码

for i in range(len(a)): if i > 0: cdf[i] += cdf[i - 1] cdf[i] = pdf[i] + cdf[i] figureIndex += 1 fig = plt.figure(figureIndex, figsize=(10, 8)) ax = fig.add_subplot(211) ax.bar(a, height=pdf, ...

Please write a genetic algorithm in Python to calculate the maximum value of function f (x)=0.4+sinc (4x)+1.1sinc (4x+2)+0.8sinc (X-2)+0.7sinc (6x-4) between -2 and 2. The function solution is represented by a 16 bit binary number. The population size is 30, the crossover probability is 0.3, the mutation probability is 0.01, and the number of iterations is 400

This is the maximum value of the `f` function that the genetic algorithm was able to find in 400 iterations with a population size of 30, a crossover probability of 0.3, and a mutation probability of ...

下面代码报错请修改一下：from pypinyin import pinyin import sys # 定义拼音汉字映射表，可以根据需要扩充 pinyin_dict = { 'a': '阿啊呵腌嗄锕', 'ai': '爱埃碍挨唉哎哀皑癌矮艾蔼隘碍捱嗳嫒霭', 'an': '安按暗岸案俺鞍氨庵胺谙埯揞犴', # ... } # 接收用户输入的拼音串 input_pinyin = input('请输入拼音串：') # 将拼音串转化为列表 pinyin_list = input_pinyin.split(' ') # 定义一个函数，用于从映射表中查找可能性最大的汉字 def find_max_probability_char(pinyin_char, candidate_chars): max_probability = 0 max_probability_char = '' for char in candidate_chars: probability = pinyin.get(pinyin_char, format='numerical')[0][0] if probability > max_probability: max_probability = probability max_probability_char = char return max_probability_char # 遍历每个拼音，查找可能性最大的汉字 result = '' for pinyin_char in pinyin_list: if pinyin_char in pinyin_dict: candidate_chars = pinyin_dict[pinyin_char] max_probability_char = find_max_probability_char(pinyin_char, candidate_chars) result += max_probability_char else: result += pinyin_char # 输出结果 print(result)

'a': '阿啊呵腌嗄锕', 'ai': '爱埃碍挨唉哎哀皑癌矮艾蔼隘碍捱嗳嫒霭', 'an': '安按暗岸案俺鞍氨庵胺谙埯揞犴', # ... } # 接收用户输入的拼音串 input_pinyin = input('请输入拼音串：') # 将拼音串转化为...

introduction to probability. 2nd ed.

### 回答1：《概率导论》是一本介绍概率理论的教材，第二版。本书旨在帮助读者理解和应用概率的基本概念和原理。在书中，作者首先介绍了概率的概念。概率是描述随机现象发生可能性的数学工具。...

Coin is heads with probability 1/3 and tails with probability 2/3. The expectation of independent flips until first heads is _____.

The expected number ...In this case, the probability of getting heads on any given flip is 1/3. Therefore, the expected number of flips until the first heads is: E(X) = 1/(1/3) = 3 So the answer is 3.

The interpretation of the quantities ma and ba is as follows. Given a Bayes-plausible measure η, the quantity ba denotes the probability that the receiver plays action a ∈ A under the optimal strategy a(·) when the sender uses the signaling scheme corresponding to η; in other words, ba � Pη(a(s¯) � a). Similarly, ma denotes the distribution of the state ω¯ , conditioned on the receiver choosing action a. By iterated expectation, we obtain ma(ω) � Eη[s¯(ω)|a(s¯) � a]. Thus, for any a ∈ A, the quantity ma denotes the mean of all posterior beliefs the receiver holds, conditioned on choosing action a. For this reason, we refer to ma as the mean posterior of the receiver corresponding to action a ∈ A. Note that ma may not correspond to any actual posterior that the receiver holds when choosing action a ∈ A; in fact, the mean posterior ma ∈ Conv(Pa) may not even lie in the set Pa. Figure 2 gives some geometric intuition for 翻译

给定一个Bayes-plausible度量η，量ba表示当发送者使用与η对应的信令方案时，接收者在最优策略a(·)下选择行动a∈A的概率；换句话说，ba表示Pη(a(s¯)∣a)。类似地，ma表示在接收者选择行动a的条件下，状态ω¯的...

prediction interval coverage probability

In other words, if we were to repeat the process of estimating a prediction interval for a new observation many times, the coverage probability would be the proportion of those intervals that ...

请修改以下代码输出正确结果不能报错：import numpy as np import matplotlib.pyplot as plt def solve_schrodinger(H): eigvals, eigvecs = np.linalg.eig(H) idx = np.argsort(eigvals) eigvals = eigvals[idx] eigvecs = eigvecs[:,idx] return eigvals, eigvecs def plot_wavefunction(x, psi): plt.plot(x, np.real(psi)) plt.plot(x, np.imag(psi)) plt.xlabel('x') plt.ylabel('psi(x)') plt.title('Wavefunction') with plt.style.context(['science', 'ieee']): plt.show() def plot_density(x, rho): plt.plot(x, rho) plt.xlabel('x') plt.ylabel('rho(x)') plt.title('Probability Density') with plt.style.context(['science', 'ieee']): plt.show() L = 10 # Box size N = 1000 # Number of grid points dx = L / N # Grid spacing x = np.linspace(-L/2, L/2, N, endpoint=False) V = np.zeros(N) # Potential energy # Kinetic energy matrix T = np.zeros((N,N)) for i in range(1,N-1): T[i,i-1] = -1/(2dx2) T[i,i] = 1/(dx2) T[i,i+1] = -1/(2dx**2) # Hamiltonian H = -0.5*T + np.diag(V) # Solve Schrodinger equation eigvals, eigvecs = solve_schrodinger(H) # Plot wavefunction of first excited state psi = eigvecs[:,1] plot_wavefunction(x, psi) # Calculate probability density of first excited state rho = np.abs(psi)**2 plot_density(x, rho)

# Calculate probability density of first excited state rho = np.abs(psi)**2 plot_density(x, rho) ``` 主要修改了以下几点： 1. `np.linalg.eig()` 改为 `np.linalg.eigh()`，因为 `H` 是实对称矩阵，eigh ...

Given a distribution P on letters, find the lowest-cost tree.

1. First, we sort the letters by their probabilities in non-increasing order. 2. Then, we create a binary tree where each leaf node represents a letter and the probability of that letter. 3. We ...

#!/usr/bin/env python # -- coding: utf-8 -- import rospy from yolov8_ros_msgs.msg import BoundingBoxes def bounding_boxes_callback(data): # 处理接收到的消息 print('Received bounding boxes:') for box in data.bounding_boxes: print(' - Class: ', box.Class) print(' Probability: ', box.probability) print(' Xmin: ', box.xmin) print(' Ymin: ', box.ymin) print(' Xmax: ', box.xmax) print(' Ymax: ', box.ymax) if name == 'main': # 初始化 ROS 节点 rospy.init_node('subscriber_node', anonymous=True) # 创建一个订阅者对象，订阅 /yolov8/BoundingBoxes 话题 rospy.Subscriber('/yolov8/BoundingBoxes', BoundingBoxes, bounding_boxes_callback) # 运行 ROS 节点 rospy.spin()这里面那一些数据可以倒入到另一个python文件中的class中的def

print(' Probability: ', box.probability) print(' Xmin: ', box.xmin) print(' Ymin: ', box.ymin) print(' Xmax: ', box.xmax) print(' Ymax: ', box.ymax) def run(self): # 运行 ROS 节点 rospy.spin...

use matlab to stimulate: For a Bernoulli trial with the success probability p, the entropy is a convex function of p

We can use MATLAB to plot the entropy function for a Bernoulli trial with success probability p and show that it is a convex function. First, we can define the entropy function as: ```matlab entropy...

用tensorflow probability写一个用于多标签分类的贝叶斯卷积神经网络

2. “A Bayesian Approach to Multi-label Classification with Joint Embeddings” by Q. Li, T. Cao, and D. Tao 3. “Bayesian Convolutional Neural Networks with Bernoulli Approximate Variational Inference...

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概率论基础教程：第8版

A first course in probability

A first course in probability(概率论基础教程)

A first course in probability 8th Edith

AttributeError: module 'tensorflow_probability.python.bijectors.bijector' has no attribute 'AutoCompositeTensorBijector'

Excess probability

Error in nomogram(data = df, fun = function(x) 0.5, varname = c("original.firstorder.RobustMeanAbsoluteDeviation", : argument "fit" is missing, with no default

introduction to probability. 2nd ed.

Coin is heads with probability 1/3 and tails with probability 2/3. The expectation of independent flips until first heads is _____.

prediction interval coverage probability

Given a distribution P on letters, find the lowest-cost tree.

use matlab to stimulate: For a Bernoulli trial with the success probability p, the entropy is a convex function of p

用tensorflow probability写一个用于多标签分类的贝叶斯卷积神经网络

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