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Chapter 15: Problem 4

Let \(\mathrm{S}=\\{1,2, \ldots ., 20\\}\). A subset \(\mathrm{B}\) of \(\mathrm{S}\) is said to be "nice", if the sum of the elements of \(\mathrm{B}\) is 203 . Than the probability that a randomly chosen subset of \(\mathrm{S}\) is "nice" is : (a) \(\frac{7}{2^{20}}\) (b) \(\frac{5}{2^{20}}\) (c) \(\frac{4}{2^{20}}\) (d) \(\frac{6}{2^{20}}\)

Short Answer

Expert verified
(b) \(\frac{5}{2^{20}}\)

Step by step solution

01

Understand the Problem

We need to determine the number of subsets of the set \( \mathrm{S}=\{1,2, \ldots, 20\} \) whose elements sum up to 203. Then, we will calculate the probability of choosing such a subset randomly from all possible subsets of \( \mathrm{S} \).
02

Calculate Total Sum of Set S

Compute the sum of all elements from 1 to 20 using the formula for the sum of an arithmetic series: \( \text{Sum} = \frac{n(n+1)}{2} \), where \( n = 20 \). Thus, \( \text{Sum} = \frac{20 \times 21}{2} = 210 \).
03

Find Sum of Elements in Set S \ B

The problem states that the sum of elements in subset \( \mathrm{B} \) is 203. Therefore, the elements not included in \( \mathrm{B} \), i.e., set \( \mathrm{S} \setminus \mathrm{B} \), must sum to \( 210 - 203 = 7 \).
04

Identify Subsets of S Having Sum 7

Determine the subsets of \( \mathrm{S} \) whose elements sum up to 7. These can be computed by evaluating all combinations of elements from 1 to 20 that add up to 7. The valid subsets are: \( \{7\}, \{1, 6\}, \{2, 5\}, \{3, 4\}, \{1, 2, 4\}\). Thus, there are 5 such subsets.
05

Calculate Total Number of Subsets of S

The total number of subsets of a set with \( n \) elements is \( 2^n \). For \( \mathrm{S} \) with 20 elements, the total number of subsets is \( 2^{20} \).
06

Calculate Probability of Selecting a "Nice" Subset

The probability of a randomly chosen subset being "nice" is the ratio of the number of "nice" subsets to the total number of subsets. Since there are 5 nice subsets, the probability is \( \frac{5}{2^{20}} \).

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Key Concepts

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

Probability
To solve problems involving probability, especially in the context of subsets, we need to understand the likelihood of selecting a particular subset with desired properties from a larger set. Probability is calculated as the ratio of favorable outcomes to the total number of possible outcomes.
  • Favorable outcomes: In this problem, these are the subsets of the set \( S \) that sum to 203.
  • Total outcomes: This is the total number of subsets of \( S \), which is calculated as \( 2^{20} \).
The probability formula in this context becomes:\[P(\text{nice subset}) = \frac{\text{Number of nice subsets}}{\text{Total number of subsets}}.\] This strategy helps in evaluating scenarios where the outcome space is large, as is the case with large sets and their subsets.
Subsets
A subset is any group that can be formed using elements from a larger set, where none, some, or all elements may be chosen. When discussing subsets, especially in problems involving summations, it's important to identify subsets with particular properties.
Key details to understand:
  • The total number of subsets of a set with \( n \) elements is \( 2^n \). This includes the empty set and the set itself.
  • Identifying subsets with specific conditions (like a particular sum) requires evaluating combinations of elements to meet the given condition.
In our context, the problem asks us to find subsets whose elements sum to 203. This entails systematic checking of combinations to ensure all possibilities are accounted for, utilizing techniques from combinatorics.
Arithmetic Series
In this problem, understanding arithmetic series provides a foundational step for calculating the total sum of a sequentially ordered set. An arithmetic series is a sequence of numbers in which the difference of any two consecutive terms is constant.
The sum formula for the first \( n \) natural numbers is used here:\[\text{Sum} = \frac{n(n+1)}{2}\]Where \( n \) is the highest number in the set. For \( S = \{1, 2, \ldots, 20\} \), \( n = 20 \), yielding total sum as 210.
This sum helps determine how any excluded subsets (\( S \setminus B \)) contribute to reaching the desired sum for subset \( B \), as it provides a baseline from which to calculate needed values.
Combinatorics
Combinatorics is a branch of mathematics dealing with counting, combination, and permutation of sets. In this exercise, combinatorial techniques are essential to identify possible subsets fitting specific criteria such as a predefined sum.
  • Applying combinatorics: Determine combinations of numbers that sum to a certain value, like 7 in this exercise, in the complement of the desired subset sum.
  • Tools: Utilizing systematic trial and error or algorithms (like dynamic programming or backtracking) helps in listing such combinations efficiently.
Understanding basic principles of combinatorics equips you with the ability to decipher how many ways elements in a set can be grouped, critical for solving complex probability and set-based puzzles.

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Most popular questions from this chapter

For three events \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), \(\mathrm{P}\) (Exactly one of \(\mathrm{A}\) or \(\mathrm{B}\) occurs) \(=\mathrm{P}(\) Exactly one of \(\mathrm{B}\) or \(\mathrm{C}\) occurs \()\) \(=\mathrm{P}\) (Exactly one of \(\mathrm{C}\) or A occurs) \(=\frac{1}{4}\) and \(\mathrm{P}\) (All the three events occur simultaneously) \(=\frac{1}{16}\). Then the probability that at least one of the events occurs, is : (a) \(\frac{3}{16}\) (b) \(\frac{7}{32}\) (c) \(\frac{7}{16}\) (d) \(\frac{7}{64}\)

\(A\) and \(B\) are events such that \(P(A \cup B)=3 / 4, P(A \cap B)=1 / 4\), \(P(\bar{A})=2 / 3\) then \(P(\bar{A} \cap B)\) is \(\quad\) [2002] (a) \(5 / 12\) (b) \(3 / 8\) (c) \(5 / 8\) (d) \(1 / 4\)

If six students, including two particular students \(A\) and \(B\), stand in a row, then the probability that \(A\) and \(B\) are separated with one student in between them is [Online May 19, 2012] (a) \(\frac{8}{15}\) (b) \(\frac{4}{15}\) (c) \(\frac{2}{15}\) (d) \(\frac{1}{15}\)

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is [2003] (a) \(\frac{2}{5}\) (b) \(\frac{4}{5}\) (c) \(\frac{3}{5}\) (d) \(\frac{1}{5}\)

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is : (a) \(\frac{21}{220}\) (b) \(\frac{3}{11}\) (c) \(\frac{1}{11}\) (d) \(\frac{2}{23}\)
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