5.4 Basic Probability, Permutations, Combinations, and Statistics

Key Takeaways

  • Theoretical probability measures favorable outcomes divided by total sample space outcomes (P = n(E)/n(S)), bounded within [0, 1], with complementary probability defining P(not E) = 1 - P(E).
  • Compound probability utilizes the multiplication rule: independent events multiply unconditional probabilities (P(A and B) = P(A)*P(B)), while dependent events without replacement adjust sample sizes on subsequent draws (P(A and B) = P(A)*P(B|A)).
  • Combinatorics distinguishes between ordered arrangements (Permutations: nPr = n! / (n-r)!, where rank, position, or order matters) and unordered selections (Combinations: nCr = n! / [r!(n-r)!], where order is irrelevant).
  • Measures of central tendency summarize distributions: Mean is the arithmetic total divided by count (sum(x)/n), Median is the middle value in ranked data (average of middle two values for even n), and Mode is the most frequently occurring score.
  • Statistical dispersion is quantified by Range (Max - Min); extreme outliers significantly distort the arithmetic mean while leaving the median unaffected.
Last updated: August 2026

5.4 Basic Probability, Permutations, Combinations, and Statistics

Core Principle: Probability, discrete combinatorics, and descriptive statistics items appear consistently across the CAT-ASVAB Mathematics Knowledge subtest. You must rapidly compute fractional probability sample spaces, distinguish permutations from combinations, calculate weighted statistical averages, and interpret graphical charts well inside the roughly 124 seconds per item that Mathematics Knowledge allows.


1. Theoretical Probability & Complementary Events

Probability measures the likelihood of an event $E$ occurring within a finite sample space $S$ of equally likely outcomes:

P(E)=Number of Favorable Outcomes n(E)Total Number of Possible Outcomes n(S)P(E) = \frac{\text{Number of Favorable Outcomes } n(E)}{\text{Total Number of Possible Outcomes } n(S)}

Core Probability Rules:

  • Probability Interval: $0 \le P(E) \le 1$ ($P = 0$ indicates an impossible event; $P = 1$ indicates absolute certainty).
  • Complementary Probability: The probability that event $E$ does not occur: P(not E)=1P(E)P(\text{not } E) = 1 - P(E)
  • "At Least One" Rule: Useful in compound trials where direct computation is lengthy: P(at least one)=1P(none)P(\text{at least one}) = 1 - P(\text{none})

Odds vs. Probability

  • Odds in Favor of $E$: $\text{Favorable Outcomes} : \text{Unfavorable Outcomes} = n(E) : [n(S) - n(E)]$
  • Odds Against $E$: $\text{Unfavorable Outcomes} : \text{Favorable Outcomes} = [n(S) - n(E)] : n(E)$
  • Example: If $P(E) = \frac{3}{8}$, the odds in favor are $3 : (8 - 3) = 3 : 5$.

2. Compound Probability: Independent vs. Dependent Events

                      COMPOUND PROBABILITY: P(A and B)
                                     |
             +-----------------------+-----------------------+
             |                                               |
      INDEPENDENT EVENTS                              DEPENDENT EVENTS
 (Occurrence of A does not affect B)             (Occurrence of A changes sample space of B)
   P(A and B) = P(A) × P(B)                       P(A and B) = P(A) × P(B | A)
  (e.g., Coin flips, replacement)               (e.g., Drawing cards without replacement)

1. Independent Events (With Replacement)

Events are independent when the outcome of the first event has no mathematical effect on the probability of subsequent events.

  • Example: Tossing a coin (Heads) AND rolling a 5 on a standard six-sided die: P(Heads and 5)=P(Heads)×P(5)=12×16=112P(\text{Heads and } 5) = P(\text{Heads}) \times P(5) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}

2. Dependent Events (Without Replacement)

Events are dependent when the outcome of the first event modifies the remaining pool of items in the sample space.

  • Example: A tactical chest contains 7 smoke grenades and 3 fragmentation grenades (total 10). Two grenades are drawn at random without replacement. What is the probability that both are smoke grenades? P(1st Smoke)=710P(\text{1st Smoke}) = \frac{7}{10} After removing 1 smoke grenade, 6 smoke and 9 total remain: P(2nd Smoke)=69=23\text{After removing 1 smoke grenade, 6 smoke and 9 total remain: } P(\text{2nd Smoke}) = \frac{6}{9} = \frac{2}{3} P(Both Smoke)=710×69=4290=715P(\text{Both Smoke}) = \frac{7}{10} \times \frac{6}{9} = \frac{42}{90} = \frac{7}{15}

3. Addition Rules: "OR" Probability

  • Mutually Exclusive (Disjoint) Events: $P(A \text{ or } B) = P(A) + P(B)$
  • Overlapping (Non-Mutually Exclusive) Events: P(A or B)=P(A)+P(B)P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)

3. Counting Principles, Factorials, Permutations & Combinations

The Fundamental Counting Principle

If action 1 can occur in $n_1$ ways, action 2 in $n_2$ ways, and action $k$ in $n_k$ ways, the total number of sequential outcomes is: Total Outcomes=n1×n2×n3××nk\text{Total Outcomes} = n_1 \times n_2 \times n_3 \times \dots \times n_k

Factorial Notation ($n!$)

A factorial is the product of all positive integers descending from $n$ to 1: n!=n×(n1)×(n2)××2×1n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1

  • $0! = 1$ (by definition), $1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5,040$

Permutations vs. Combinations Decision Matrix

+-----------------------------------------------------------------------------------------+
|                              PERMUTATIONS VS. COMBINATIONS                              |
+------------------------------------+----------------------------------------------------+
| PERMUTATION (Order MATTERS)        | COMBINATION (Order DOES NOT Matter)                |
| - Arrangements, Rankings, Roles    | - Groups, Teams, Committees, Selections            |
| - Order: 1st, 2nd, 3rd / PIN code  | - Group: {Soldier A, Soldier B} has no order       |
|                                    |                                                    |
|             n!                     |                   n!                               |
|     nPr = ──────                 |          nCr = ──────────────                          |
|            (n - r)!                |                r!(n - r)!                          |
+------------------------------------+----------------------------------------------------+

Rapid Non-Calculator Computation Shortcuts:

  • Permutation Shortcut ($_{7}P_3$): Start at 7 and multiply 3 descending factors: 7P3=7×6×5=210{}_{7}P_3 = 7 \times 6 \times 5 = 210
  • Combination Shortcut ($_{8}C_3$): Multiply 3 descending factors in numerator, divided by $3!$ in denominator: 8C3=8×7×63×2×1=3366=56{}_{8}C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

4. Measures of Central Tendency & Statistical Spread

+-----------------------------------------------------------------------------------------+
|                               4 CORE STATISTICAL MEASURES                               |
+-----------------------+-----------------------------------------------------------------+
| 1. Mean (Average)     | Sum of all values divided by total count: x̄ = (Σ x) / n         |
| 2. Median (Middle)    | Central value when data is ordered ascendingly                  |
| 3. Mode (Frequency)   | Most frequently occurring value (unimodal, bimodal, or none)    |
| 4. Range (Spread)     | Difference between maximum and minimum value: Max - Min         |
+-----------------------+-----------------------------------------------------------------+

Calculating the Median

Always arrange the data from least to greatest first:

  • Odd Sample Size ($n = 7$): ${4, 7, 8, \mathbf{11}, 15, 18, 22} \implies \text{Median} = 11$ (exact 4th element).
  • Even Sample Size ($n = 6$): ${3, 6, \mathbf{8, 12}, 15, 19} \implies \text{Median} = \frac{8 + 12}{2} = \frac{20}{2} = 10$.

The "Target Mean" Problem Technique

A frequent ASVAB question asks: "A recruit earns scores of 80, 84, and 92 on three exams. What score is required on the 4th exam to achieve an average of 88?"

  • Total Points Method: Target Sum=4×88=352\text{Target Sum} = 4 \times 88 = 352 Current Sum=80+84+92=256\text{Current Sum} = 80 + 84 + 92 = 256 Required Score=352256=96\text{Required Score} = 352 - 256 = 96

Weighted Averages

When data points carry different credit weights or percentages: xˉw=(wixi)wi\bar{x}_w = \frac{\sum (w_i \cdot x_i)}{\sum w_i} Example: An assessment weights lab work at 30% (score 85) and the final test at 70% (score 95): xˉw=(0.30×85)+(0.70×95)=25.5+66.5=92\bar{x}_w = (0.30 \times 85) + (0.70 \times 95) = 25.5 + 66.5 = 92

Outlier Sensitivity

  • Mean: Highly sensitive to extreme outlier values (pulled toward extremes).
  • Median: Robust and resistant to outliers (ideal measure for skewed distributions).

5. Interpreting Graphical & Tabular Data

  • Bar Graphs: Compare discrete quantities across categories. Always verify the vertical axis baseline to detect compressed scales.
  • Line Graphs: Show trends over time. The steepness of the line segment represents the rate of change.
  • Pie Charts (Circle Graphs): Represent parts of a whole ($100% = 360^\circ$). To convert a percentage to a central sector angle: Central Angle θ=(Percentage100)×360\text{Central Angle } \theta = \left(\frac{\text{Percentage}}{100}\right) \times 360^\circ (Example: $25% \rightarrow 0.25 \times 360^\circ = 90^\circ$; $15% \rightarrow 0.15 \times 360^\circ = 54^\circ$)
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Permutations vs Combinations Decision Flowchart
Test Your Knowledge

An ammunition container holds 6 armor-piercing rounds, 4 tracer rounds, and 2 standard rounds. If two rounds are selected at random without replacement, what is the probability that both selected rounds are tracer rounds?

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Test Your Knowledge

A military commander must select a 3-person specialized reconnaissance patrol from a platoon squad of 8 qualified soldiers. How many different reconnaissance patrol teams can be formed?

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Test Your Knowledge

A student earns scores of 76, 84, 89, and 81 on four module exams. What score must the student achieve on the fifth module exam to attain an overall average mean score of exactly 84?

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Test Your Knowledge

In a physical fitness battalion competition with 7 participants, trophies are awarded for 1st place (Gold), 2nd place (Silver), and 3rd place (Bronze). In how many distinct ways can the three trophies be awarded?

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