Home Page Checker Bonus Question: Compound Random experiment: Flip a penny, draw a marble from a bag containg 3 red, 5 orange, and 2 green marbles. Draw the tree diagram to represent this compound random experiment. List the sample space. Compute the probabilities for each of the atomic events of the sample space. Event A = { Red marble and head on penny or green marble tail on penny) Event B = { orange or green marble and head or tail on the penny) Calculate: p(A), p(B), p(A union B), and finally determine if they are independent events.

Home Page Checker Bonus Question: Compound Random experiment: Flip a penny, draw a marble from a bag containg 3 red, 5 orange, and 2 green marbles. Draw the tree diagram to represent this compound random experiment. List the sample space. Compute the probabilities for each of the atomic events of the sample space. Event A = { Red marble and head on penny or green marble tail on penny) Event B = { orange or green marble and head or tail on the penny) Calculate: p(A), p(B), p(A union B), and finally determine if they are independent events.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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def solve_n_queens(n): def is_safe(board, row, col): # Verificar la columna for i in range(row): if board[i][col] == 1: return False # Verificar la diagonal izquierda i, j = row, col while i >= 0 and j >= 0: if board[i][j] == 1: return False i -= 1 j -= 1 # Verificar la diagonal derecha i, j = row, col while i >= 0 and j < n: if board[i][j] == 1: return False i -= 1 j += 1 return True def solve(board, row): if row >= n: solution = [] for r in board: solution.append("".join(['R' if x == 1 else '.' for x in r])) solutions.append(solution) return for col in range(n): if is_safe(board, row, col): board[row][col] = 1 solve(board, row + 1)…
def solve_n_queens(n): def is_safe(board, row, col): # Verificar la columna for i in range(row): if board[i][col] == 1: return False # Verificar la diagonal izquierda i, j = row, col while i >= 0 and j >= 0: if board[i][j] == 1: return False i -= 1 j -= 1 # Verificar la diagonal derecha i, j = row, col while i >= 0 and j < n: if board[i][j] == 1: return False i -= 1 j += 1 return True def solve(board, row): if row >= n: solution = [] for r in board: solution.append("".join(['Q' if x == 1 else '.' for x in r])) solutions.append(solution) return for col in range(n): if is_safe(board, row, col): board[row][col] = 1 solve(board, row + 1)…
: |: Section 3 - Random Forests Decision Trees on their own are effectiev classifiers. The true power, though, comes from a forest of trees -- multiple decision trees working together as an ensemble learner. We create multiple trees, each using a subset of the available attributes, let them each make a guess at the correct classification, and then take the majority vote as the predicted class. Sounds tricky, right? Once again, sklearn to the rescue. We are going to see if a random forest can improve on our decision tree accuracy for the bank data. To begin, we are going to create a Random Forest using the wheat training and test data from the previous section. First, reload our data. 1 # reload the wheat dataset from UCI 2 3 df = pd. read_csv ("seeds_dataset.txt", sep='\\t', engine='python') 4 6 7 df.columns = ['a', 'p', 'compactness', 'length', 'width', 'coeff', 'length_g', 'type'] print (f'Our data has {df.shape [0]} rows and {df.shape [1]} columns') 8 #Mark 70% of the data for…