Textbook Exercises & Solutions

Given

Rank and explain: (i) next Monday comes after Sunday; (ii) snow in Mumbai in July; (iii) an elephant walks through your classroom today; (iv) greeting at least one friend at school tomorrow.

To Find

Qualitative likelihoods on the impossible-to-certain scale

Reasoning

Contextual likelihood is not a measured numerical probability.

Solution

1. (i) Certain, 1: this follows from the defined order of weekdays.

2. (ii) Extremely unlikely in Mumbai’s ordinary July climate

   no measured probability is supplied, so do not manufacture one or infer logical impossibility merely from rarity.

3. (iii) Extremely unlikely in an ordinary classroom

   the answer depends on the situation.

   It is not a logical impossibility in every imaginable setting.

4. (iv) More likely if tomorrow is a normal school day and friends attend

   holidays, absence and other circumstances can change the judgement.

Result

Certain for (i); contextual, reasoned judgements for (ii)–(iv). No exact probabilities are justified for those three.

Given

A random sample of 30 sweets contains 10 red, 8 green, 7 yellow and 5 blue. (i) Find P(green) when selecting from this sample. (ii) Estimate yellow sweets in a bag of 600.

To Find

Sample probability and population estimate

Reasoning

Uniform selection from the sample; sample proportion used as an estimate for the bag.

Solution

1. (i) P(green) = $\frac{8}{30}$ = $\frac{4}{15}$.

2. (ii) Estimated yellow count = 600 × $\frac{7}{30}$ = 140.

   A sample estimate need not equal the actual bag count.

Result

(i) $\frac{4}{15}$. (ii) An estimated 140 yellow sweets.

Given

Among 40 sampled students, 14 prefer Science, 11 Arts, 9 Sports and 6 Debate. The school has 800 students. Find sample P(Arts) and estimate the whole-school Sports count.

To Find

Arts sample probability and estimated Sports count

Reasoning

Use the sampled proportion, keeping estimate and actual count distinct.

Solution

1. P(Arts) = $\frac{11}{40}$.

2. Estimated Sports count = 800 × $\frac{9}{40}$ = 180.

Result

$\frac{11}{40}$; an estimated 180 students.

Given

Toss a coin 20 times. Record heads and tails, calculate experimental P(heads), and consider P(tails) on the next toss.

To Find

Observed counts, relative frequency and next-toss model

Reasoning

Experimental probabilities use actual observations; theoretical probabilities need a model.

Solution

1. Let the recorded head count be h.

   Then the tail count is 20−h.

   Enter your actual counts

   there is no predetermined observation.

2. Experimental P(heads) = $\frac{h}{20}$.

3. If the coin is fair and tosses are independent, the next-toss P(tails) = $\frac{1}{2}$, regardless of previous counts.

   Without that model, the 20 observations alone do not establish an exact next-toss probability.

Result

Head count h, tail count 20−h, observed head proportion $\frac{h}{20}$; next-tail probability $\frac{1}{2}$ under the fair-independent model.

Given

Toss a paper cup 100 times and record bottom, top and side landings. Assign experimental probabilities.

To Find

Three observed landing proportions

Reasoning

Relative frequency = observed count / total trials.

Solution

1. Let b,t,s be your recorded counts.

   Check b+t+s = 100.

2. Assign $\frac{b}{100}$, $\frac{t}{100}$ and $\frac{s}{100}$ to bottom, top and side respectively.

3. These estimates add to 1.

   Three named outcomes do not imply probabilities of $\frac{1}{3}$: the cup’s shapes and landing dynamics differ.

Result

Use your actual $\frac{b}{100}$, $\frac{t}{100}$ and $\frac{s}{100}$; no landing counts are fabricated.

Given

Find P(an even number) for a fair six-sided die.

To Find

Probability of an even outcome

Reasoning

Each of six faces is equally likely.

Solution

1. S = {1,2,3,4,5,6}

   E = {2,4,6}.

2. P(E) = $\frac{3}{6}$ = $\frac{1}{2}$.

Result

$\frac{1}{2}$.

Given

A die shows 3 in three of 12 rolls. Find experimental and theoretical P(3), explain the difference, and discuss 60, 600 and 6000 rolls.

To Find

Observed proportion, theoretical probability and long-run interpretation

Reasoning

Relative frequency and the fair-die theoretical model are different quantities.

Solution

1. Observed P(3) = $\frac{3}{12}$ = $\frac{1}{4}$.

   Under a fair-die model, theoretical P(3) = $\frac{1}{6}$.

2. Random variation makes a small sample differ from the model probability.

3. For 60, 600 and 6000 independent fair rolls, the expected counts are 10, 100 and 1000

   these are not guaranteed counts.

4. Relative frequency tends to stabilise near $\frac{1}{6}$ over many trials.

   It need not equal $\frac{1}{6}$ exactly or move closer after every added roll.

Result

Observed $\frac{1}{4}$; theoretical $\frac{1}{6}$ if fair; longer runs typically give a more stable relative frequency, not exact equality.

Given

How many outcomes are in the sample space of one six-sided die roll?

To Find

Sample-space size

Reasoning

List each possible face once.

Solution

1. S = {1,2,3,4,5,6}, so n(S) = 6.

Result

6.

Given

Write S for (i) a die and coin together, (ii) a random integer between −5 and +5, (iii) drawing one ball from 5 green and 7 red balls. Parts (ii)–(iii) are on page 168.

To Find

Complete sample spaces with outcomes identified

Reasoning

Choose what is being recorded and distinguish physical objects when counting equally likely outcomes.

Solution

1. (i) S = {$\left(1,H\right)$,$\left(1,T\right)$,$\left(2,H\right)$,$\left(2,T\right)$,$\left(3,H\right)$,$\left(3,T\right)$,$\left(4,H\right)$,$\left(4,T\right)$,$\left(5,H\right)$,$\left(5,T\right)$,$\left(6,H\right)$,$\left(6,T\right)$}.

2. (ii) Including the endpoints: S = {−5,−4,−3,−2,−1,0,1,2,3,4,5}.

   If “between” is intended strictly, exclude −5 and 5

   the wording does not specify inclusion.

3. (iii) For recorded colour, S = {green,red}

   these are not equally likely.

   For uniformly selected individual balls, label them: S = {G1,G2,G3,G4,G5,R1,R2,R3,R4,R5,R6,R7}.

Result

(i) 12 ordered outcomes; (ii) 11 integers inclusively or 9 strictly; (iii) two colour outcomes or 12 individually labelled balls.

Given

Choose a snack from Samosa, Pakora, Bhaji and a drink from Chai, Lassi. List S and the event of choosing Samosa.

To Find

All combinations and the Samosa event

Reasoning

Use ordered pairs (snack,drink).

Solution

1. S = {(Samosa,Chai),(Samosa,Lassi),(Pakora,Chai),(Pakora,Lassi),(Bhaji,Chai),(Bhaji,Lassi)}.

2. E = {(Samosa,Chai),(Samosa,Lassi)}.

   No equal-likelihood claim is needed just to list outcomes.

Result

Six combinations; two belong to the Samosa event.

Given

Basket A has one apple and two oranges; basket B has one banana and one mango. Pick one fruit randomly from each. Give a tree, sample space and P(apple and banana).

To Find

Fruit-pair outcomes and apple–banana probability

Reasoning

Select each physical fruit uniformly; selections from separate baskets are independent.

Solution

1. Label the oranges O1,O2.

   The tree has first branches Apple, O1, O2, each of probability $\frac{1}{3}$.

   Each splits into Banana and Mango, each with probability $\frac{1}{2}$.

2. S = {(Apple,Banana),(Apple,Mango),(O1,Banana),(O1,Mango),(O2,Banana),(O2,Mango)}.

   The six labelled pairs are equally likely.

3. P(Apple,Banana) = $\frac{1}{3}$ × $\frac{1}{2}$ = $\frac{1}{6}$.

   If oranges are grouped by type, the four type-pairs are not equally likely.

Result

P(apple and banana) = $\frac{1}{6}$.

Given

A box has 3 red, 4 black and 2 green pens. You select a pen uniformly and replace it; your friend then selects. Give the colour tree and P(same colour).

To Find

Colour-pair sample space and same-colour probability

Reasoning

Replacement restores the original counts for the second selection.

Solution

1. First branches: R = $\frac{3}{9}$, B = $\frac{4}{9}$, G = $\frac{2}{9}$.

   From each branch the second branches again have R = $\frac{3}{9}$, B = $\frac{4}{9}$, G = $\frac{2}{9}$.

2. Colour pairs: RR,RB,RG,BR,BB,BG,GR,GB,GG.

   Their probabilities respectively are $\frac{9}{81}$,$\frac{12}{81}$,$\frac{6}{81}$,$\frac{12}{81}$,$\frac{16}{81}$,$\frac{8}{81}$,$\frac{6}{81}$,$\frac{8}{81}$,$\frac{4}{81}$.

3. P(same) = P(RR)+P(BB)+P(GG) = $\frac{\left(9+16+4\right)}{81}$ = $\frac{29}{81}$.

Result

$\frac{29}{81}$. The nine colour pairs are not equally likely.

Given

Fill the blanks: (i) impossible-event probability; (ii) name of the set of all outcomes; (iii) certain-event probability; (iv) fair-coin P(heads). Parts (ii)–(iv) continue on page 170.

To Find

The four missing entries

Reasoning

Definitions of probability, certainty and sample space.

Solution

1. Impossible events have probability 0

   all outcomes form the sample space.

2. A certain event has probability 1

   a fair coin assigns $\frac{1}{2}$ to heads.

Result

(i) 0; (ii) sample space; (iii) 1; (iv) $\frac{1}{2}$.

Given

15 of 50 surveyed students like football. Identify the frequency and relative frequency.

To Find

Count and proportion for football preference

Reasoning

Relative frequency divides the frequency by the sample size.

Solution

1. The frequency is 15 students.

2. Relative frequency = $\frac{15}{50}$ = $\frac{3}{10}$ = 0.3.

   The source blank requesting a fraction/decimal is therefore “relative frequency”.

Result

Frequency 15; relative frequency $\frac{3}{10}$ or 0.3.

Given

Are outcomes equally likely for (i) a car starting/not starting, (ii) a fair coin, (iii) a fair die, (iv) a random marble from 3 red and 7 blue, (v) a baby being a boy or girl?

To Find

Which stated outcomes justify equal likelihood

Reasoning

Two named possibilities need not each have probability $\frac{1}{2}$.

Solution

1. (i) Not established: vehicle condition affects starting probability.

2. (ii) Yes: fair means P(H) = P(T) = $\frac{1}{2}$.

   (iii) Yes: each fair-die face has probability $\frac{1}{6}$.

3. (iv) Colour outcomes are unequal: $\frac{3}{10}$ red, $\frac{7}{10}$ blue.

   Individual marbles can be equally likely when uniformly drawn.

4. (v) Exact equal likelihood is not established merely by naming two outcomes.

   A population model or evidence is needed

   do not infer an exact $\frac{1}{2}$ from the wording alone.

Result

Equal likelihood is given for (ii) and (iii); it is not justified for the other stated outcome groupings.

Given

Give S and the requested probability: (i) at least one head with two coins; (ii) even card from 1–10; (iii) die greater than 4; (iv) not red in a bag of 3 red,2 blue,1 green; (v) exactly two heads with three coins.

To Find

Five sample spaces and event probabilities

Reasoning

For coin and die calculations assume fair independent trials; select individual cards/balls uniformly.

Solution

1. (i) S = {HH,HT,TH,TT}

   E = {HH,HT,TH}

   P = $\frac{3}{4}$.

2. (ii) S = {1,2,3,4,5,6,7,8,9,10}

   E = {2,4,6,8,10}

   P = $\frac{5}{10}$ = $\frac{1}{2}$.

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

   E = {5,6}

   P = $\frac{2}{6}$ = $\frac{1}{3}$.

4. (iv) S = {R1,R2,R3,B1,B2,G1}

   E = {B1,B2,G1}

   P = $\frac{3}{6}$ = $\frac{1}{2}$.

5. (v) S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

   E = {HHT,HTH,THH}

   P = $\frac{3}{8}$.

Result

(i) $\frac{3}{4}$; (ii) $\frac{1}{2}$; (iii) $\frac{1}{3}$; (iv) $\frac{1}{2}$; (v) $\frac{3}{8}$, under the stated models.

Given

Select one of three candies—strawberry, lemon and mint—uniformly at random.

To Find

Probability of selecting strawberry

Reasoning

One favourable candy among three equally selectable candies.

Solution

1. S = {strawberry,lemon,mint}.

   The strawberry event has one outcome.

2. P = $\frac{1}{3}$.

Result

$\frac{1}{3}$.

Given

Choose a red or blue shirt and jeans, khakis or shorts. List all outfits in a table.

To Find

All shirt–pants combinations

Reasoning

For each of two shirts there are three pants choices.

Solution

1. Red shirt: jeans

   khakis

   shorts.

2. Blue shirt: jeans

   khakis

   shorts.

3. Every row choice of shirt can pair with every column choice of pants, so there are 2 × 3 = 6 outfits.

Result

(Red,jeans), (Red,khakis), (Red,shorts), (Blue,jeans), (Blue,khakis), (Blue,shorts).

Given

Among 1000 tyres, recorded counts are: below 4000 km:20; 4001–9000:210; 9001–14000:325; above 14000:445. Estimate P(i) below 4000, (ii) between 4000 and 14000, (iii) above 14000.

To Find

Three probabilities from recorded frequencies

Reasoning

Use the supplied table bins as recorded; these are empirical estimates.

Solution

1. (i) $\frac{20}{1000}$ = $\frac{1}{50}$.

2. (ii) Combining the two middle bins: $\frac{\left(210+325\right)}{1000}$ = $\frac{535}{1000}$ = $\frac{107}{200}$.

3. (iii) $\frac{445}{1000}$ = $\frac{89}{200}$.

4. The printed bins leave the exact 4000 boundary unstated.

   The middle-bin answer follows the table’s grouping

   no extra observation at that boundary is invented.

Result

(i) $\frac{1}{50}$; (ii) $\frac{107}{200}$ from the middle bins; (iii) $\frac{89}{200}$.

Given

Choose one of the five PEACE letter cards uniformly. Find (i) P, E or C, and (ii) not E.

To Find

Probabilities for the specified letter events

Reasoning

Distinguish the two physical E cards.

Solution

1. S = {P,E1,A,C,E2}, with five equally likely cards.

2. For the event “P or E or C”, favourable cards are P,E1,C,E2, giving $\frac{4}{5}$.

   If separate letter probabilities are requested, P(P) = $\frac{1}{5}$, P(E) = $\frac{2}{5}$ and P(C) = $\frac{1}{5}$.

3. Not E comprises P,A,C, so P(not E) = $\frac{3}{5}$ = 1$\frac{-2}{5}$.

Result

(i) Union event $\frac{4}{5}$; individual P,E,C probabilities $\frac{1}{5}$,$\frac{2}{5}$,$\frac{1}{5}$. (ii) $\frac{3}{5}$.