Number and Numeration Systems: Set Theory and Probability Solutions

Application of Set Theory in Examination Analysis

In the study of Number and Numeration Systems, set theory provides a robust framework for solving complex word problems involving groups and their overlaps. Question 9 of the examination material focuses on a group of candidates who passed two specific subjects: Integrated Science and Mathematics. The total number of candidates who passed either Integrated Science or Mathematics is given as $60$. Mathematically, let $S$ represent the set of candidates who passed Integrated Science and $M$ represent the set of candidates who passed Mathematics. The union of these two sets, denoted as $n(S \cup M)$, is equal to $60$.

Furthermore, the transcript identifies that $15$ candidates passed both subjects. This intersection of the two sets is represented as $n(S \cap M) = 15$. To identify the individual set sizes, the problem provides a relative comparison: 9 more candidates passed Mathematics than Integrated Science. If we define the number of candidates who passed Integrated Science as $n(S) = x$, then the number of candidates who passed Mathematics is defined as $n(M) = x + 9$.

Detailed Calculation of Subject Pass Rates

To find the number of candidates who passed in each individual subject, we utilize the Principle of Inclusion-Exclusion for two sets. The formula is expressed as:

$n(S \cup M) = n(S) + n(M) - n(S \cap M)$

By substituting the known values and algebraic expressions into this formula, we get:

$60 = x + (x + 9) - 15$

To solve for $x$, we first simplify the right side of the equation:

$60 = 2x - 6$

Adding $6$ to both sides of the equation yields:

$66 = 2x$

Dividing both sides by $2$ results in:

$x = 33$

Consequently, the number of candidates who passed Integrated Science is $n(S) = 33$. Using the relationship established earlier, the number of candidates who passed Mathematics is calculated as follows:

$n(M) = 33 + 9 = 42$

Therefore, $33$ candidates passed Integrated Science and $42$ candidates passed Mathematics.

Probability Distribution and Venn Diagram Analysis

Part (b) of the problem requires finding the probability that a candidate passed exactly one subject. To determine this, we must first calculate the number of candidates who passed only Integrated Science and the number of candidates who passed only Mathematics, excluding those who passed both.

The number of candidates who passed only Integrated Science is calculated by subtracting the intersection from the total set for Science:

$n(S \cap M^c) = n(S) - n(S \cap M)$

$33 - 15 = 18$

The number of candidates who passed only Mathematics is calculated by subtracting the intersection from the total set for Mathematics:

$n(M \cap S^c) = n(M) - n(S \cap M)$

$42 - 15 = 27$

The total number of candidates who passed exactly one subject is the sum of these two subsets:

$18 + 27 = 45$

To find the probability, we divide the number of candidates who passed exactly one subject by the total number of candidates in the union (those who passed at least one subject):

$P(\text{Exactly One Subject}) = \frac{45}{60}$

This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is $15$:

$\frac{45 \div 15}{60 \div 15} = \frac{3}{4}$

In decimal form, this probability is expressed as $0.75$.

Introduction to Survey Statistics and Quality Control

Question 10 introduces a new scenario involving a survey of products. Specifically, the transcript mentions a survey conducted on $100$ bulbs. This initial figure represents the size of the universal set or the total sample population for the subsequent statistical or set-based analysis concerning the bulbs' attributes or performance metrics. Although the content for Question 10 is truncated, it establishes a framework for analyzing large datasets in numerical systems.