Comprehensive Guide to Probability Rules, Combinations, and the Binomial Distribution

Fundamentals of Probability

  • Probability Range and Definition:

    • Probability is measured on a scale from 0.000.00 to 1.001.00.

    • A probability of 0.000.00 indicates that an event is impossible.

    • A probability of 1.001.00 indicates that an event is certain.

  • Classical Probability Formulas:

    • If an event can occur in AA ways and fail to occur in BB ways, and all outcomes are equally likely:

      • Probability of the event occurring: P=AA+BP = \frac{A}{A + B}

      • Probability of the event not occurring: P=BA+BP = \frac{B}{A + B}

  • Event Relationships:

    • Mutually Exclusive Events: The occurrence of one event precludes the occurrence of another event (for example, pregnancy or not pregnancy).

    • Independent Events: The occurrence of one event has no effect on the probability of the occurrence of another event (for example, consecutive coin flips).

  • Operational Decision Clues:

    • "OR": Indicates that to determine the probability of either of two events occurring, their probabilities are added.

    • "AND": Indicates that to determine the probability of two events occurring together or in succession, their probabilities are multiplied.

The Additive Rule of Probability

  • Additive Rule for Mutually Exclusive Events:

    • For events that cannot happen simultaneously, the probability of occurrence of one event or another equals the sum of their individual probabilities.

    • General formula: P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)

    • Example (Multipack of Crisps):

      • A multipack contains 1010 bags total: 33 Cheese & Onion (C&O), 33 Ready Salted (RS), and 44 Pickled Onion (PO).

      • Probability of selecting a C&O or a PO bag:

        • P(C&O or PO)=P(C&O)+P(PO)P(\text{C\&O or PO}) = P(\text{C\&O}) + P(\text{PO})

        • P(C&O or PO)=310+410=710=0.7P(\text{C\&O or PO}) = \frac{3}{10} + \frac{4}{10} = \frac{7}{10} = 0.7

  • Additive Rule for Non-Mutually Exclusive Events:

    • When two events can occur simultaneously, their joint occurrence is counted twice in a simple sum; therefore, the joint probability must be subtracted.

    • General formula: 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)

    • Example (Rolling a Single Six-Sided Die):

      • Problem: Determine the probability of rolling a number greater than 33 or an even number on a single 6-sided die.

      • Numbers greater than 33: {4,5,6}\{4, 5, 6\}, giving P(greater than 3)=36P(\text{greater than } 3) = \frac{3}{6}

      • Even numbers: {2,4,6}\{2, 4, 6\}, giving P(even)=36P(\text{even}) = \frac{3}{6}

      • Numbers that are simultaneously greater than 33 and even: {4,6}\{4, 6\}, giving P(greater than 3 and even)=26P(\text{greater than } 3 \text{ and even}) = \frac{2}{6}

      • Calculation:

        • P(A or B)=36+36−26=46=23P(A \text{ or } B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}

The Multiplicative Rule and Conditional Probability

  • Multiplicative Rule for Independent Events:

    • The probability of the joint occurrence of two or more independent events is the product of their separate individual probabilities.

    • General formula: P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)

    • Example (Selection With Replacement):

      • A multipack contains 1010 bags: 33 C&O, 33 Ready Salted, and 44 Pickled Onion.

      • Two bags are selected, replacing the first bag before selecting the second.

      • Probability of selecting a C&O and then another C&O:

        • P(C&O)=310=0.3P(\text{C\&O}) = \frac{3}{10} = 0.3

        • P(C&O and C&O)=P(C&O)×P(C&O)P(\text{C\&O and C\&O}) = P(\text{C\&O}) \times P(\text{C\&O})

        • P(C&O and C&O)=0.3×0.3=0.09P(\text{C\&O and C\&O}) = 0.3 \times 0.3 = 0.09

  • Multiplicative Rule for Dependent Events (Conditional Probability):

    • Conditional Probability: Defined as the probability of an event given the prior occurrence of another event, denoted as P(B∣A)P(B|A).

    • Notation detail: The vertical bar denotes "under the condition", not division or a fraction.

    • General formula: P(A and B)=P(A)×P(B∣A)P(A \text{ and } B) = P(A) \times P(B|A)

    • Example 1 (Selection Without Replacement - Crisps):

      • Multipack contains 1010 bags: 33 C&O, 33 Ready Salted, and 44 Pickled Onion.

      • Two bags are selected without replacement.

      • Probability of choosing a C&O followed by another C&O:

        • P(2 C&O)=P(C&O)×P(C&O given first C&O)P(2\text{ C\&O}) = P(\text{C\&O}) \times P(\text{C\&O given first C\&O})

        • P(A)=310P(A) = \frac{3}{10}

        • P(B∣A)=29P(B|A) = \frac{2}{9}

        • Calculation: 310×29=690≈0.07\frac{3}{10} \times \frac{2}{9} = \frac{6}{90} \approx 0.07

    • Example 2 (Card Draw Without Replacement):

      • A card is chosen at random from a standard deck of 5252 playing cards. Without replacement, a second card is chosen.

      • Probability that the first card is a queen and the second card is a jack:

        • P(queen on first pick)=452P(\text{queen on first pick}) = \frac{4}{52}

        • P(jack on 2nd pick given queen on 1st pick)=451P(\text{jack on 2nd pick given queen on 1st pick}) = \frac{4}{51}

        • Joint probability calculation:

          • P(queen and jack)=452×451=162652=4663≈0.006P(\text{queen and jack}) = \frac{4}{52} \times \frac{4}{51} = \frac{16}{2652} = \frac{4}{663} \approx 0.006

Combinations and Factorials

  • Definition of Combinations:

    • Combinations measure the number of ways to arrange or select a subset of objects where order does not matter.

  • Factorial Definition:

    • Indicated by an exclamation mark (!!).

    • Defined as multiplying a given integer by all smaller positive integers (for example, 5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120).

  • Computational Formula for Combinations:


    Formula for combinations NCr

    • Formula expression: NCr=N!r!(N−r)!NCr = \frac{N!}{r!(N - r)!}

    • Variables:

      • NN = Number of events

      • rr = Number of successes

  • Worked Example (A-Level Exam Outcomes):

    • Problem: Determine how many ways a student can achieve 33 A grades out of 55 A-Level exams.

    • Parameters: N=5N = 5, r=3r = 3

    • Evaluation:

      • NCr=5!3!(5−3)!NCr = \frac{5!}{3!(5 - 3)!}

      • NCr=5!3!(2!)=1206×2=12012NCr = \frac{5!}{3!(2!)} = \frac{120}{6 \times 2} = \frac{120}{12}

      • NCr=10NCr = 10

The Binomial Distribution

  • Purpose and Structure:

    • Calculates both combinations and the specific probability for any one combination of getting rr successes out of NN events.

    • Formula representation:


      Formula for binomial probability distribution

      • Mathematical formula: P(r)=N!r!(N−r)!pr(1−p)N−rP(r) = \frac{N!}{r!(N - r)!} p^r (1 - p)^{N - r}

  • Variable Definitions:

    • P(r)P(r) = Probability of exactly rr successes

    • NN = Number of events / trials

    • rr = Number of successes

    • pp = Probability of success on any one event / trial

    • 1−p1 - p = Probability of failure on any one event / trial

  • Large Sample Approximation:

    • When the number of events (NN) is large, the binomial distribution is equal to the normal distribution.

  • Worked Example (Guessing on Multiple-Choice Test):

    • Problem: Determine the probability that an individual with no knowledge gets exactly 33 correct answers on a 55-question multiple-choice test (MCQ), where each question has 77 possible answers and only one correct answer.

    • Parameters:

      • N=5N = 5

      • r=3r = 3

      • Probability of success on any one event: p=17≈0.14p = \frac{1}{7} \approx 0.14

      • Probability of failure on any one event: 1−p=1−17=67≈0.861 - p = 1 - \frac{1}{7} = \frac{6}{7} \approx 0.86

    • Step-by-step evaluation:

      • Combinations factor: NCr=5!3!(5−3)!=10NCr = \frac{5!}{3!(5 - 3)!} = 10

      • Single sequence probability factor: pr(1−p)N−r=(17)3(67)5−3p^r (1 - p)^{N - r} = \left(\frac{1}{7}\right)^3 \left(\frac{6}{7}\right)^{5 - 3}

      • Full substitution: P(3)=[5!3!(5−3)!](17)3(67)5−3P(3) = \left[\frac{5!}{3!(5 - 3)!}\right] \left(\frac{1}{7}\right)^3 \left(\frac{6}{7}\right)^{5 - 3}

      • Decimal approximation: P(3)=(10)(0.14)3(0.86)2P(3) = (10)(0.14)^3 (0.86)^2

      • Intermediate computation: P(3)=(10)(0.003)(0.74)P(3) = (10)(0.003)(0.74)

      • Final result: P(3)=0.022P(3) = 0.022 (2%2\% )

Practice Problems

  • Question 1:

    • A student completely guesses on a 66-question multiple-choice test. Each question has 44 possible answers, with only one correct. Using the binomial distribution, what is the probability they get exactly 22 correct?

  • Question 2:

    • A student guesses on a 1010-question multiple-choice test. Each question has 55 possible answers, with only one correct. What is the probability of getting exactly 44 correct?

  • Question 3:

    • A player takes 88 free throws in a game and makes each one with probability 0.70.70.70.7. What is the probability they make exactly 66?

  • Question 4:

    • A coin is flipped 77 times. What is the probability of getting exactly 55 heads?

  • Question 5 (A bit trickier):

    • A student guesses on a 1212-question test. Each question has 33 possible answers, with only one correct. What is the probability of getting exactly 22 correct?