12.2 Basic Statistics, Permutations, Combinations & Probability

Basic Statistics, Permutations, Combinations & Probability

Statistical analysis, counting principles, and probability theory represent fundamental quantitative skills assessed on the NTS GAT General examination. Whether calculating the combined average score of student cohorts, evaluating arrangement possibilities for a panel, or determining the likelihood of compound events, mastering these mathematical frameworks provides a distinct competitive advantage. This section provides a thorough treatment of measures of central tendency, weighted statistics, combinatorics (permutations and combinations), and classical probability rules.

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1. Statistical Measures of Central Tendency & Dispersion

A. Arithmetic Mean (Average)

The arithmetic mean ($\bar{x}$) represents the central measure of a numerical dataset containing $n$ elements: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{x_1 + x_2 + \dots + x_n}{n}$

• The Fundamental Sum Identity: $\text{Sum of Values} = \bar{x} \times n$. On the NTS GAT, when an additional element is added or removed from a group, always convert given means into total sums first.

B. Weighted Average and Combined Mean

When individual data groups have unequal sizes ($n_1, n_2, \dots, n_k$) or varying weights ($w_1, w_2, \dots, w_k$), calculating a simple average of the group means produces an incorrect result. The Weighted Mean ($\bar{x}w$) and Combined Mean ($\bar{x}{\text{comb}}$) must be used: $\bar{x}_w = \frac{\sum (w_i x_i)}{\sum w_i}$ $\bar{x}_{\text{comb}} = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2 + \dots + n_k \bar{x}_k}{n_1 + n_2 + \dots + n_k}$

C. Median

The median is the middle observation in an ordered dataset arranged in ascending or descending sequence.

• Odd Number of Observations ($n$): The median is the single value located at position $\frac{n+1}{2}$.
• Even Number of Observations ($n$): The median is the arithmetic average of the two middle values located at positions $\frac{n}{2}$ and $\frac{n}{2} + 1$.
• Property: The median is resistant to extreme outliers, whereas the mean is sensitive to extreme values.

D. Mode and Range

• Mode: The data value occurring with the highest frequency. A dataset may be unimodal (one mode), bimodal (two modes), multimodal, or have no mode (if all values occur with equal frequency).
• Range: A basic measure of dispersion calculated as the difference between the maximum and minimum observations: $\text{Range} = x_{\text{max}} - x_{\text{min}}$.

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2. Fundamentals of Combinatorics: Permutations & Combinations

Combinatorics provides counting techniques to calculate the total number of possible arrangements or selections without manually listing every outcome.

A. Fundamental Counting Principle

If a primary operation can be performed in $m$ distinct ways, and a secondary independent operation can be performed in $n$ distinct ways, then both operations in sequence can be performed in $m \times n$ total ways.

B. Factorial Notation

For any positive integer $n$: $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$ (By mathematical convention, $0! = 1$).

C. Permutations ($nPr$) — Order Matters

A permutation is an ordered arrangement of $r$ elements chosen from a set of $n$ distinct elements. Use permutations when positions, titles, sequences, or ranks are distinct: $nPr = \frac{n!}{(n-r)!}$

• Permutations with Repetition: The number of distinct arrangements of $n$ items where $p$ items are identical of type 1, $q$ items are identical of type 2, etc., is: $\text{Arrangements} = \frac{n!}{p! \times q! \times r!}$

D. Combinations ($nCr$) — Order Does Not Matter

A combination is a selection of $r$ elements chosen from a set of $n$ distinct elements where the order of selection is irrelevant (e.g., forming a committee, selecting a team): $nCr = \frac{n!}{r!(n-r)!} = \frac{nPr}{r!}$

• Key Symmetry Property: $nCr = nC(n-r)$ (e.g., $10C7 = 10C3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$).

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3. Probability Theory & Rules of Events

Probability measures the mathematical likelihood of an event $E$ occurring within a sample space $S$ containing all possible equally likely outcomes: $P(E) = \frac{\text{Number of Favorable Outcomes } n(E)}{\text{Total Number of Possible Outcomes } n(S)}$

• Probability Range: $0 \le P(E) \le 1$.

A. Complementary Probability Rule

The probability that event $E$ does not occur ($E'$) is: $P(E') = 1 - P(E)$

• GAT Exam Shortcut ("At Least One" Rule): When asked to calculate the probability of getting "at least one" successful outcome across multiple trials, calculate the complement (getting zero successes) and subtract from 1: $P(\text{At least 1 Success}) = 1 - P(\text{Zero Successes})$

B. Addition Rule for Mutually Exclusive and Non-Exclusive Events

• Mutually Exclusive Events: Events $A$ and $B$ cannot happen simultaneously ($P(A \cap B) = 0$). $P(A \text{ or } B) = P(A \cup B) = P(A) + P(B)$
• Non-Mutually Exclusive Events: Events $A$ and $B$ can overlap. $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

C. Multiplication Rule for Independent Events

Two events $A$ and $B$ are independent if the occurrence of one does not alter the probability of the other: $P(A \text{ and } B) = P(A \cap B) = P(A) \times P(B)$

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4. Step-by-Step Worked Mathematical Examples

Worked Example 1: Combined Mean Calculation

Scenario: A GAT preparation class consists of three sections. Section A has 20 students with an average score of 75. Section B has 30 students with an average score of 85. Section C has 50 students with an average score of 90.

Question: What is the combined mean score for all 100 students across the three sections?

• Step 1: Calculate total sum of marks for Section A: $\text{Sum}_A = 20 \times 75 = 1,500$.
• Step 2: Calculate total sum of marks for Section B: $\text{Sum}_B = 30 \times 85 = 2,550$.
• Step 3: Calculate total sum of marks for Section C: $\text{Sum}_C = 50 \times 90 = 4,500$.
• Step 4: Sum total marks and total students: $\text{Total Marks} = 1,500 + 2,550 + 4,500 = 8,550$ $\text{Total Students} = 20 + 30 + 50 = 100$
• Step 5: Apply combined mean formula: $\bar{x}_{\text{comb}} = \frac{8,550}{100} = 85.5$

Worked Example 2: Combinations for Committee Formation

Scenario: A university department consists of 6 male professors and 5 female professors. A research committee of 4 members is to be formed.

Question: In how many ways can the committee be formed such that it contains exactly 2 male professors and 2 female professors?

• Step 1: Calculate ways to select 2 male professors from 6: $6C2 = \frac{6 \times 5}{2 \times 1} = 15$.
• Step 2: Calculate ways to select 2 female professors from 5: $5C2 = \frac{5 \times 4}{2 \times 1} = 10$.
• Step 3: Apply Fundamental Counting Principle (multiply options): $\text{Total Ways} = 6C2 \times 5C2 = 15 \times 10 = 150 \text{ ways}$

Worked Example 3: Probability ("At Least One" Coin Toss)

Scenario: A fair six-sided die is rolled 3 times consecutively.

Question: What is the probability of rolling at least one '6' across the 3 rolls?

• Step 1: Identify probability of NOT rolling a '6' on a single roll: $P(\text{Not } 6) = \frac{5}{6}$.
• Step 2: Calculate probability of rolling NO '6's in all 3 independent rolls: $P(\text{No } 6 \text{ in 3 rolls}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$
• Step 3: Apply complementary probability rule: $P(\text{At least one } 6) = 1 - \frac{125}{216} = \frac{216 - 125}{216} = \frac{91}{216}$

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5. Summary Table: Permutation vs. Combination Decision Matrix