The Mathematics of Maybe: Introduction to Probability

(i) Monday follows Sunday in the weekly cycle, so this is certain: probability 1.

(ii) Snow in Mumbai in July is treated as impossible in the usual classroom model; an everyday judgement is extremely unlikely.

(iii) An elephant entering an ordinary classroom is extremely unlikely; calling it impossible assumes circumstances that exclude it.

(iv) Greeting a friend is usually more likely than not if you attend school and have friends there. Its precise probability needs information about your circumstances.

The sample has 30 sweets, including 8 green and 7 yellow. Estimate each probability by its sample fraction.

P(green) ≈ 8/30 = 4/15. In 600 sweets the estimated yellow count is 600 × 7/30.

There are 40 sampled students. Arts has 11 and sports has 9.

P(arts) ≈ 11/40. Estimated sports membership among 800 is 800 × 9/40.

Perform 20 tosses and record H heads and T tails, with H + T = 20.

The experimental probabilities are H/20 and T/20. They depend on your actual observations.

For a fair independent next toss, the theoretical probability of tails stays 1/2, whatever happened earlier.

Record bottom, top and side counts b, t and s over 100 drops. Check b + t + s = 100.

The estimates are b/100, t/100 and s/100. Three named outcomes do not imply equal probabilities: the cup’s shape affects how it lands.

For a fair six-sided die the outcomes 1–6 are equally likely.

The favourable faces are 2, 4 and 6.

Three occurrences of 3 in twelve rolls give 3/12 = 1/4 experimentally.

A fair die gives P(3) = 1/6 theoretically. More rolls usually make the relative frequency a better estimate, but no finite set of rolls must match 1/6 exactly.

List every possible number on the top face of a standard six-sided die.

(i) A die and a coin give S = {(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)}.

(ii) Interpreting the endpoints as included, S = {−5,−4,−3,−2,−1,0,1,2,3,4,5}. If “between” excludes endpoints, omit −5 and 5.

(iii) For the colour observation, S = {green, red}. With 5 green and 7 red balls, these colour outcomes have probabilities 5/12 and 7/12; they are not equally likely. Individual labelled balls form an equally likely 12-outcome space.

Let the three snacks be S1, S2 and S3, with S1 denoting samosa, and the two drinks D1 and D2.

List (S1,D1), (S1,D2), (S2,D1), (S2,D2), (S3,D1), (S3,D2).

The two combinations containing samosa are (S1,D1) and (S1,D2).

First bag: apple has probability 1/3 and orange 2/3. Second bag: banana and mango each have probability 1/2.

Multiply probabilities along a branch: AB = 1/6, AM = 1/6, OB = 1/3, OM = 1/3.

For an equally likely space, distinguish the two oranges as O1 and O2; the six outcomes are AB, AM, O1B, O1M, O2B, O2M.

The colour probabilities are red 3/9, black 4/9 and green 2/9. Replacement makes these probabilities unchanged for the second draw.

The nine ordered colour pairs are RR, RK, RG, KR, KK, KG, GR, GK, GG (K denotes black). Their probabilities are 9, 12, 6, 12, 16, 8, 6, 8, 4 divided by 81.

For matching colours, add RR, KK and GG: (9 + 16 + 4)/81.

An impossible event has probability 0. All possible outcomes together form the sample space.

A certain event has probability 1. A fair coin gives either face probability 1/2.

The event occurred 15 times in 50 trials.

Relative frequency is occurrences divided by trials, 15/50.

(i) A car starting or failing to start gives two outcomes, but they need not have equal chances.

(ii) A fair coin has equally likely faces. (iii) A fair die has equally likely faces.

(iv) Three balls of one colour and seven of another give probabilities 3/10 and 7/10, so colour outcomes are unequal.

(v) “Boy or girl” is often treated as an approximately equal two-outcome classroom model. Equality here is an assumption, not a consequence of having two labels.

(i) Two fair coins have sample space {HH,HT,TH,TT}; at least one head occurs in HH,HT,TH, giving 3/4.

(ii) S={1,2,3,4,5,6,7,8,9,10}; the five even labels give 1/2. (iii) S={1,2,3,4,5,6}; faces 5 and 6 give 1/3.

(iv) Label the individual balls R1,R2,R3,B1,B2,G1: three of six are not red, so 1/2.

(v) S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}. Exactly two heads occur in HHT,HTH,THH, so 3/8.

Count strawberry choices among all equally likely choices in the question.

The strawberry fraction is one out of three.

Pair each of the two shirts with each of the three pairs of pants.

There are 2 × 3 = 6 outfits. Labels S1, S2 and P1, P2, P3 preserve the textbook’s items without using its artwork.

There are 1000 tyres in total. The four recorded bin counts are 20, 210, 325 and 445.

Use the recorded bins: the first interval gives 20/1000; the middle two together give (210 + 325)/1000; the final interval gives 445/1000.

These are estimates from observations. Use the table’s interval conventions at the boundary lifetimes.

The five letter positions are equally likely. Distinguish the two Es as E1 and E2: S={P,E1,A,C,E2}.

The event P, E or C contains P,E1,C,E2: four of five positions. Not E contains P,A,C: three of five.

Each of eight equal sectors has probability 1/8.

Favourable labels: {8}; {1,3,5,7}; {3,4,5,6,7,8}; all eight labels; {3,6}.

Initially there are 4 red and 5 blue balls. After the first draw only 8 remain.

Red then blue: (4/9)(5/8) = 5/18. Blue then blue: (5/9)(4/8) = 5/18.

Two ordinary dice can produce sums from 2 to 12.

A sum of 13 is impossible. The event “sum lies from 2 through 12” includes every outcome and is certain.

The sample space consists of 36 ordered die pairs (a,b), with each coordinate from 1 to 6.

The relevant prime sums are 7 and 11. Sum 7 has six pairs; sum 11 has two.

There are 4 red, 3 green and 2 blue balls: 9 altogether. Without replacement there are 9 × 8 = 72 equally likely ordered ball pairs.

Same-colour ordered pairs total 4×3 + 3×2 + 2×1 = 20. Different-colour pairs total 72 − 20 = 52.

Three fair tosses give eight equally likely ordered outcomes.

HHT and HTH satisfy both conditions. THH has two heads but fails the first-head condition.

Permute the digits 1, 2, 3, 4 without repetition: 4! = 24 numbers.

An even number ends in 2 or 4. For either last digit, the first three digits have 3! arrangements.

Each question has one correct option out of four, so a guess is correct with probability 1/4.

Choose which one of the three questions is wrong: 3 choices. Multiply 3 × (1/4)² × (3/4).

With replacement: S = {(i,j): i,j ∈ {1,2,3,4}}, giving 16 ordered outcomes.

Without replacement remove (1,1),(2,2),(3,3),(4,4), leaving 12 outcomes.

Pair H and T separately with each card 1–6.

S = {(H,1),(H,2),(H,3),(H,4),(H,5),(H,6),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}.

In three tosses the head count can be 0, 1, 2 or 3, including the all-tail and all-head outcomes.

For fair independent coins the respective probabilities are 1/8, 3/8, 3/8 and 1/8. The four counts are not equally likely.

List (i) omits 0; list (ii) omits 3; list (iii) includes the impossible count 4. Only list (iv) includes exactly all possible counts.

Assume the drop is equally likely per unit area across the 3 × 2 rectangle. Its area is 6 square units.

The circle has diameter 1, hence radius 1/2 and area π/4. Divide its area by the rectangle’s area.

A fair unpredictable toss gives either team the same probability 1/2 of winning the choice.

Fairness also requires choosing a side before the result is known and using the same agreed procedure for both teams.

Your friend can guess a possible result but cannot be certain of an unrevealed random toss.

A correct guess once does not establish that future tosses are predictable.

Independent rolls of a fair die retain the same six equiprobable faces.

Earlier fours do not make a 4 “due” or prevent it occurring again.

A sample space must include every possible outcome relevant to the observation.

When counting favourable outcomes over total outcomes, verify equal likelihood. Distinguish individual balls or repeated letters when their multiplicities matter.

Two independent fair tosses have outcomes HH, HT, TH and TT.

HT and TH are two different ordered ways to get one of each.