Probability & Counting Principles (Sections 9.1–9.4)

Section 9.1 – Determining Probabilities

  • Key vocabulary

    • Experiment = an activity that produces observable results (outcomes).

    • Outcome = each individual result of an experiment.

    • Sample space SS = set of all possible outcomes; outcomes in SS are mutually exclusive & collectively exhaustive.

    • Event = any subset of SS.

  • Modeling a sample space

    • Can be shown by sketches, tree-diagrams, or roster form {\,}.

    • Example: single coin toss → S=H,TS={H,T}.

    • Die-rolling example

    • Sample space: S=1,2,3,4,5,6S={1,2,3,4,5,6} (drawing, tree diagram, set list).

    • Events defined

      • A=5A={5} (rolling a 5)

      • B=2,4,6B={2,4,6} (even number)

      • C=2,3,5C={2,3,5} (prime number)

Experimental (Empirical) Probability

  • Based on data from actually performing the experiment.

  • Expressed as fraction/decimal/percent of number of times event occurs over total trials.

  • Accuracy increases as number of trials grows → Law of Large Numbers / Bernoulli’s Theorem (Theorem 9-1)

    • “As the number of trials nn increases, the experimental probability P^(A)\hat P(A) approaches the theoretical probability P(A)P(A).”

Theoretical Probability (Equally Likely Outcomes)

  • If all outcomes in SS are equally likely:

    P(A)=n(A)n(S)P(A)=\frac{n(A)}{n(S)}

  • Single fair coin → P(H)=P(T)=12P(H)=P(T)=\tfrac12.

  • Theorem 9-2: For any event AA, 0P(A)10\le P(A)\le1.

Worked example (Choosing a random integer 1!!251!\text{–}!25)

  • Given S=1,2,,25S={1,2,\dots,25}, all equally likely.

    • Even number (A): n(A)=12n(A)=12P(A)=1225P(A)=\tfrac{12}{25}.

    • Number > 25 (B): n(B)=0n(B)=0P(B)=0P(B)=0.

    • Number < 26 (C): n(C)=25n(C)=25P(C)=1P(C)=1.

    • Prime (D): n(D)=9n(D)=9P(D)=925P(D)=\tfrac{9}{25}.

    • Even & prime (E): n(E)=1n(E)=1 (only 2) → P(E)=125P(E)=\tfrac{1}{25}.


Section 9.2 – Rules for Combined Events

Mutually Exclusive (Disjoint) Events

  • Definition: AB=A \cap B = \varnothing (cannot occur together).

  • Additive rule (Theorem 9-4) for mutually exclusive: P(AB)=P(A)+P(B).P(A\cup B)=P(A)+P(B).

General Addition Rule

P(AB)=P(A)+P(B)P(AB).P(A\cup B)=P(A)+P(B)-P(A\cap B).

Independent Events

  • Definition: Occurrence of AA has no influence on BBP(BA)=P(B).P(B|A)=P(B).

  • Multiplicative rule (Theorem 9-7): P(AB)=P(A)P(B).P(A\cap B)=P(A)\,P(B).

Dice Illustrations

  • Rolling two fair dice (36 equally likely ordered pairs).

    • Event A: “double sixes” ⇒ 1/361/36.

    • Event B: “sum 7 or 11” ⇒ 836=29\tfrac{8}{36}=\tfrac{2}{9}.

    • A & B share no common outcome ⇒ mutually exclusive.

  • Sum even (E) vs sum prime (F)

    • n(E)=18,  n(F)=15,  n(EF)=1n(E)=18,\; n(F)=15,\; n(E\cap F)=1 (only sum 2).

    • P(EF)=1836+1536136=3236=89.P(E\cup F)=\tfrac{18}{36}+\tfrac{15}{36}-\tfrac{1}{36}=\tfrac{32}{36}=\tfrac{8}{9}.

Sequential Sampling – With vs. Without Replacement

  • Box with 11 letters, specific word “BABY” drawn in order.

    • Without replacement:

      P(BABY)=2111102918=13960.P(BABY)=\frac{2}{11}\cdot\frac{1}{10}\cdot\frac{2}{9}\cdot\frac{1}{8}=\frac{1}{3\,960}.

    • With replacement:

      P(BABY)=(211)2(111)2=414641.P(BABY)=\left(\frac{2}{11}\right)^2\left(\frac{1}{11}\right)^2=\frac{4}{14\,641}.

    • Tree diagrams illustrate successive conditional probabilities.


Section 9.3 – Applications: Odds & Expected Value

Odds

  • For event AA with probability P(A)P(A):

    • Odds in favor =P(A)1P(A)=\dfrac{P(A)}{1-P(A)} → expressed as m:nm:n where m=P(A),  n=1P(A)m=P(A),\;n=1-P(A) after scaling to integers.

    • Odds against =1P(A)P(A)=\dfrac{1-P(A)}{P(A)} (reverse ratio).

  • Equally likely outcomes version:

    • Odds in favor =\dfrac{\text{# favorable}}{\text{# unfavorable}}.

  • Examples

    • Die, number < 5 ⇒ favorable 4, unfavorable 2 → odds 2:1.

    • Fair coin, heads ⇒ odds 1:1.

    • Ace from 52-card deck ⇒ odds 1:12.

    • Heart from deck ⇒ odds 1:3.

  • Theorem 9-8: If odds in favor are m:nm:n then P(E)=mm+nP(E)=\dfrac{m}{m+n}; if odds against are m:nm:n then P(E)=nm+n.P(E)=\dfrac{n}{m+n}.

Expected Value (Mathematical Expectation)

E=<em>i=1na</em>iPi.E=\sum<em>{i=1}^{n} a</em>i\,P_i.

  • Fair game ⇒ (expected payoff) − (cost to play) =0.=0.

  • Coin-toss game example

    • Payoffs: 1010 (HH), 55 (exactly one H), 00 (TT).

    • Probabilities: 14,  12,  14.\tfrac14,\;\tfrac12,\;\tfrac14.

    • E=10(14)+5(12)+0(14)=5E=10\left(\tfrac14\right)+5\left(\tfrac12\right)+0\left(\tfrac14\right)=5.

    • Cost =5=5 ⇒ net =0=0 → fair.

Dice-Sum Game (cost 55)

  • Payoff structure summarized:

    • Sum 2 or 12 → 1010 (prob 1+136=236\tfrac{1+1}{36}=\tfrac{2}{36}).

    • Sum 3 or 11 → 99 (prob 236\tfrac{2}{36}).

    • Sum 4 or 10 → 88 (prob 3+336=636\tfrac{3+3}{36}=\tfrac{6}{36}).

    • Sum 5 or 9 → 55 (prob 4+436=836\tfrac{4+4}{36}=\tfrac{8}{36}).

    • Sum 6,7,8 → 00 (prob 5+6+536=1636\tfrac{5+6+5}{36}=\tfrac{16}{36}).

  • Expected winnings (before cost): E=10(236)+9(236)+8(636)+5(836)+0(1636)=4E=10\left(\tfrac{2}{36}\right)+9\left(\tfrac{2}{36}\right)+8\left(\tfrac{6}{36}\right)+5\left(\tfrac{8}{36}\right)+0\left(\tfrac{16}{36}\right)=4.

  • Net expectation =45=1=4-5=-1 → unfair; favors the operator; average player loss $1 per game → rational decision: do not play.


Section 9.4 – Counting Techniques

Permutations (Ordered Arrangements, No Repetition)

  • Number of permutations of nn objects taken rr at a time:

    nPr=n!(nr)!.nP_r=\frac{n!}{(n-r)!}. (Order matters.)

  • Illustrations

    • All 9 actors arranged in a line: 9P9=9!=362880.9P_9=9!=362\,880.

    • Choosing 3 initials (no repeats) from 26 letters: 26P3=26×25×24=15600.26P_3=26\times25\times24=15\,600.

Combinations (Unordered Selections)

  • Number of combinations of nn objects taken rr at a time:

    (nr)=nC<em>r=n!r!(nr)!=nP</em>rr!.\binom{n}{r}=nC<em>r=\frac{n!}{r!(n-r)!}=\frac{nP</em>r}{r!}.

  • Logic: count permutations then divide by r!r! arrangements of each chosen group.

  • Illustrations

    • Book-club chooses 3 of 42 books: (423)=11480.\binom{42}{3}=11\,480.

    • Handshakes among 25 students (choose any 2): (252)=300.\binom{25}{2}=300.

Mixed Combination/Probability Example – Class Committee

  • Class: 12 girls, 10 boys (total 22).

    a) Number of committees of 5 with 3 girls & 2 boys:

    (123)(102)=9900.\binom{12}{3}\binom{10}{2}=9\,900.

    b) Probability random 5-member committee has 3 G + 2 B:

    (123)(102)(225)=990026334=501330.376.\dfrac{\binom{12}{3}\binom{10}{2}}{\binom{22}{5}}=\dfrac{9\,900}{26\,334}=\tfrac{50}{133}\approx0.376.

    c) Committees with no boys: (100)(125)=792.\binom{10}{0}\binom{12}{5}=792.

    d) Probability of all-girl committee: 79226334=41330.030.\dfrac{792}{26\,334}=\tfrac{4}{133}\approx0.030.


Summary of Key Probability Properties (from text)

  1. P()=0P(\varnothing)=0 (impossible event).

  2. P(S)=1P(S)=1 (certain event).

  3. For any event AA, 0P(A)10\le P(A)\le1.

  4. If events A,BA,B are mutually exclusive: P(AB)=P(A)+P(B).P(A\cup B)=P(A)+P(B).

  5. General addition: P(AB)=P(A)+P(B)P(AB).P(A\cup B)=P(A)+P(B)-P(A\cap B).

  6. Complement rule: P(Ac)=1P(A).P(A^c)=1-P(A).

  7. Independence: P(AB)=P(A)P(B).P(A\cap B)=P(A)P(B).

  8. Odds–Probability conversion: if odds are m:nm:n in favor, P=mm+nP=\tfrac{m}{m+n}.

These principles link experimental data, theoretical models, and counting techniques, enabling systematic analysis of random events, fair-game design, and decision-making under uncertainty.