First, list what can hap­pen

Before a cricket match the cap­tains toss a coin, and every­one agrees it is a fair way to decide. Prob­a­bil­ity starts to become pre­cise the moment you write down every out­come that could pos­si­bly hap­pen. So we begin with ran­dom exper­i­ments, sam­ple spaces and events. Then we learn to com­bine events with "and", "or", "not" and "but not", see what mutu­ally exclu­sive and exhaus­tive events are, and fin­ish with the few axioms every prob­a­bil­ity must obey.

Ran­dom exper­i­ments and sam­ple spaces

A ran­dom exper­i­ment is one that has more than one pos­si­ble out­come, so you can­not pre­dict exactly what will hap­pen. Its sam­ple space SS is the set of all pos­si­ble out­comes, and each sin­gle out­come is called a sam­ple point.

  • Toss two coins: S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}.
  • Roll a die and toss a coin: S={1H,1T,2H,…,6T}S = \{1H, 1T, 2H, \ldots, 6T\}, 1212 points.
  • Toss a coin until a head appears: S={H,TH,TTH,…}S = \{H, TH, TTH, \ldots\}, infi­nite.
  • Draw two balls from a bag of 22 red and 33 blue. You can describe SS by colours, {RR,RB,BR,BB}\{RR, RB, BR, BB\}, or by the indi­vid­ual balls, 5C2=10^5C_2 = 10 points.
Tree diagram for tossing two coins: the first coin branches to H and T, each branches again to H and T, giving the four outcomes HH, HT, TH and TT.
Toss­ing two coins: four sam­ple points
Grid of the sample space for rolling a die and tossing a coin: die faces 1 to 6 across, coin H and T down, giving the twelve points 1H to 6H and 1T to 6T.
Rolling a die and toss­ing a coin: 6 x 2 = 12 sam­ple points

Events

An event is a sub­set of SS. We say it occurs when the out­come we actu­ally get belongs to it. ∅\varnothing is the impos­si­ble event, SS is the sure event, and an event with just one point is called sim­ple.

Com­bin­ing events. A∪BA \cup B means AA or BB (or both). A∩BA \cap B means both hap­pen. A′A' (not AA) =S−A= S - A. And A−B=A∩B′A - B = A \cap B' means AA but not BB.

Mutu­ally exclu­sive: A∩B=∅A \cap B = \varnothing, so the two can never hap­pen together. Exhaus­tive: A1∪⋯∪An=SA_1 \cup \cdots \cup A_n = S, so at least one of them is bound to hap­pen.

Exam­ple 1. Roll a die and let AA = even ={2,4,6}= \{2, 4, 6\}, BB = prime ={2,3,5}= \{2, 3, 5\}, CC = at most 22 ={1,2}= \{1, 2\}. A∩B={2}A \cap B = \{2\}; A∪C={1,2,4,6}A \cup C = \{1, 2, 4, 6\}; B′={1,4,6}B' = \{1, 4, 6\}; A−B={4,6}A - B = \{4, 6\}. AA and {1,3,5}\{1, 3, 5\} are mutu­ally exclu­sive and exhaus­tive.

Venn diagram in S = {1, ..., 6}: A (even) only holds 4 and 6, the overlap A and B holds 2, B (prime) only holds 3 and 5, and 1 lies outside both.
Exam­ple 1: the events even and prime on a die

Axioms of prob­a­bil­ity

A prob­a­bil­ity PP on a finite sam­ple space gives each event a num­ber, fol­low­ing three sim­ple rules:

  1. P(E)≥0P(E) \ge 0 for every event EE;
  2. P(S)=1P(S) = 1;
  3. if EE and FF are mutu­ally exclu­sive, P(E∪F)=P(E)+P(F)P(E \cup F) = P(E) + P(F).

Every­thing else fol­lows from these. For instance P(∅)=0P(\varnothing) = 0, P(E)≤1P(E) \le 1, and P(E)P(E) is the sum of the prob­a­bil­i­ties of its sam­ple points. When all nn out­comes are equally likely, P(E)=n(E)n(S)\displaystyle P(E) = \tfrac{n(E)}{n(S)}.

Exam­ple 2. Is the assign­ment P(ω1)=0.1P(\omega_1) = 0.1, P(ω2)=0.35P(\omega_2) = 0.35, P(ω3)=0.25P(\omega_3) = 0.25, P(ω4)=0.3P(\omega_4) = 0.3 valid? Each value is non-neg­a­tive and together they add to 11, so yes. Had we used 0.40.4 in place of 0.350.35, the total would be 1.051.05, and the answer would be no.

Exam­ple 3. A die is loaded so that each even num­ber is twice as likely as each odd num­ber. Give each odd face prob­a­bil­ity pp and each even face 2p2p. Then 9p=19p = 1; P(number>3)=P(4)+P(5)+P(6)=2+1+29=59\displaystyle P(\text{number} > 3) = P(4) + P(5) + P(6) = \tfrac{2 + 1 + 2}{9} = \tfrac{5}{9}.

Try these your­self

  1. Write down the sam­ple space for each: (a) a coin tossed three times; (b) a die rolled twice (how many points?); (c) a fam­i­ly's two chil­dren by sex, older first.
  2. A coin is tossed and, if it shows a head, a die is rolled. Describe SS.
  3. Two dice are rolled. List: AA = sum is 88; BB = both show the same num­ber; CC = first shows 55. Find A∩BA \cap B, A∩CA \cap C, B∪CB \cup C in terms of counts.
  4. Which pairs from ques­tion 3 are mutu­ally exclu­sive?
  5. A card is drawn from a pack. AA = heart, BB = king, CC = red. Describe A∩BA \cap B, A∪CA \cup C, C−AC - A.
  6. Check whether P(a)=13\displaystyle P(a) = \tfrac{1}{3}, P(b)=12\displaystyle P(b) = \tfrac{1}{2}, P(c)=14\displaystyle P(c) = \tfrac{1}{4} is a valid assign­ment on {a,b,c}\{a, b, c\}.
  7. A coin is biased so that heads is three times as likely as tails. Find P(H)P(H) and P(T)P(T).
  8. In the loaded die of Exam­ple 3, find P(prime)P(\text{prime}).
  9. Give an exam­ple of three events that are mutu­ally exclu­sive but not exhaus­tive.
  10. Show that AA and A′A' are always mutu­ally exclu­sive and exhaus­tive.

Answers to check against

Show answers
  1. (a) 88 out­comes HHH, …, TTT; (b) 3636 ordered pairs; (c) {BB,BG,GB,GG}\{BB, BG, GB, GG\}.
  2. {H1,H2,…,H6,T}\{H1, H2, \ldots, H6, T\}: 77 points.
  3. A={(2,6),(3,5),(4,4),(5,3),(6,2)}A = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}; BB has 66 points; CC has 66. A∩B={(4,4)}A \cap B = \{(4,4)\}; A∩C={(5,3)}A \cap C = \{(5,3)\}; n(B∪C)=6+6−1=11n(B \cup C) = 6 + 6 - 1 = 11.
  4. None of the three pairs.
  5. King of hearts; all red cards (hearts are red); dia­monds.
  6. They add to 1312>1\displaystyle \tfrac{13}{12} > 1, so it is not valid.
  7. 34\displaystyle \tfrac{3}{4}, 14\displaystyle \tfrac{1}{4}.
  8. P(2)+P(3)+P(5)=2+1+19=49\displaystyle P(2) + P(3) + P(5) = \tfrac{2 + 1 + 1}{9} = \tfrac{4}{9}.
  9. On a die: {1},{2},{3}\{1\}, \{2\}, \{3\}.
  10. A∩A′=∅A \cap A' = \varnothing and A∪A′=SA \cup A' = S.