Chapter 24 — Probability

Many things in life are uncertain. When a farmer sows seeds he does not know exactly how many will germinate; when a batsman faces a ball he does not know whether it will be a wide, a dot ball or a six. We use words such as probably, most likely, there is a fair chance, it is almost impossible to talk about such things. These words give a rough feeling of chance, but two people can mean different things by 'probably'. Mathematics replaces these words by a single number called the probability of the event.

The word probability comes from 'probable', which means likely. The subject began in the seventeenth century when the French mathematicians Blaise Pascal and Pierre de Fermat exchanged letters about problems arising in games of dice. Later the Italian mathematician Girolamo Cardano and then Jacob Bernoulli and Pierre Simon Laplace developed the theory. Today probability is used in insurance, weather prediction, medical trials, the study of heredity, quality control in factories, election surveys and computer science.

The basic idea is simple. Suppose an action can produce several results and we cannot say in advance which one will occur, but every result has the same chance. Then the chance of a particular collection of results is the fraction of all results that belong to that collection. If a coin is tossed, there are two results, head or tail, and each is as likely as the other; so the chance of a head is one out of two, written 1/2 or 0.5 or 50 per cent.

Notice three things in this example. First, the result of a single toss cannot be predicted; the coin might show a tail even though we said the chance of a head is 1/2. Second, the number 1/2 tells us what to expect in the long run: if the coin is tossed a very large number of times, roughly half the tosses will be heads. Third, the number does not depend on who is tossing or on the past tosses; a coin has no memory. If five heads have appeared in a row, the chance of a head on the sixth toss is still 1/2.

In this chapter we shall give exact meanings to the words experiment, outcome and event, learn to count favourable and total outcomes, and then compute probabilities for coins, dice, cards, bags of balls and regions on a plane. Throughout, the answer is a fraction, decimal or percentage between 0 and 1, and the student should always check that the answer lies in this range.

An experiment in probability means any action or process whose result we observe. Tossing a coin, throwing a die, drawing a card from a pack, picking a ball from a bag, noting the sex of a new-born child, checking whether a bulb is defective, are all experiments. An experiment is called a random experiment if it satisfies two conditions: it has more than one possible result, and it is not possible to predict in advance which result will occur, though all the possible results are known.

Each possible result of a random experiment is called an outcome. When a coin is tossed the outcomes are head (H) and tail (T). When a die is thrown the outcomes are the numbers 1, 2, 3, 4, 5, 6 on the upper face. When a card is drawn from a well-shuffled pack the outcomes are the 52 cards.

The collection of all possible outcomes of a random experiment is called its sample space, usually denoted by S. Thus for a coin S = {H, T}; for a die S = {1, 2, 3, 4, 5, 6}; for a new-born child S = {boy, girl}. The number of outcomes in the sample space is written n(S).

When an experiment has two stages, we list every combination. Two coins tossed together give S = {HH, HT, TH, TT}, so n(S) = 4. Note that HT (head on the first coin, tail on the second) and TH are different outcomes, because the coins are distinct even if they look alike. Three coins give 2 × 2 × 2 = 8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Two dice thrown together give 6 × 6 = 36 outcomes, written as ordered pairs (1, 1), (1, 2), …, (6, 6). A careful list, or a 6 × 6 table, is the safest way to count them.

A single trial is one performance of the experiment; a throw of a die is one trial. Repeating a trial many times and recording the outcomes forms the basis of experimental probability, discussed later. A student should get into the habit of writing the sample space first in every problem: many mistakes in this chapter come from miscounting the total outcomes, such as thinking two dice give 12 outcomes instead of 36, or that two coins give 3 outcomes (two heads, one head, no head) instead of 4.

An event is any collection of outcomes of a random experiment; in the language of sets, an event is a subset of the sample space. When a die is thrown, 'getting an even number' is the event E = {2, 4, 6}; 'getting a number greater than 4' is the event F = {5, 6}; 'getting a 3' is the event {3}, which contains a single outcome and is called an elementary event or simple event. An event made of more than one outcome is a compound event.

We say an event has occurred if the outcome of the trial belongs to that event. If the die shows 6, the events E and F above have both occurred, but the event {3} has not. The outcomes that belong to an event are called the outcomes favourable to that event, and their number is written n(E).

The outcomes of an experiment are said to be equally likely if there is no reason to expect any one of them in preference to the others. A fair coin gives head and tail equally likely; a fair die gives each of the six numbers equally likely; a well-shuffled pack gives each card the same chance of being drawn. On the other hand, if a die is loaded by making one face heavier, or if a bag contains 5 red and 1 blue ball and we ask 'red or blue', the outcomes red and blue are not equally likely.

The word 'fair', 'unbiased' or 'well-shuffled' in a problem is the signal that the outcomes are equally likely and the classical definition may be applied. When outcomes are not equally likely we must first break them into equally likely pieces. In the bag with 5 red and 1 blue ball the six individual balls are equally likely, and then the event 'red' has 5 favourable outcomes out of 6.

It is important to describe events precisely. 'A number less than 3' on a die is {1, 2}; 'a number at most 3' is {1, 2, 3}; 'a prime number' is {2, 3, 5}; 'a multiple of 3' is {3, 6}. Reading the phrase carefully and writing the set of favourable outcomes explicitly is the second habit of a careful student, after writing the sample space. The probability is then simply n(E) divided by n(S).

Let a random experiment have n(S) equally likely outcomes and let E be an event with n(E) favourable outcomes. The theoretical probability or classical probability of E is defined as

P(E) = n(E) / n(S) = (number of outcomes favourable to E) / (total number of equally likely outcomes).

This definition was given by Laplace and is the one used throughout this chapter. Because n(E) is a count of some of the outcomes and n(S) is the count of all of them, n(E) is never negative and never more than n(S). Dividing, we get the fundamental fact 0 ≤ P(E) ≤ 1. The probability is often left as a fraction in lowest terms, but it may also be written as a decimal or a percentage.

Consider a die. The probability of getting a 4 is 1/6. The probability of an even number is 3/6 = 1/2. The probability of a number greater than 2 is 4/6 = 2/3. The probability of a number less than 7 is 6/6 = 1, and the probability of the number 7 is 0/6 = 0.

The definition applies only when the outcomes are equally likely. A common error is to say 'a student either passes or fails, so the probability of passing is 1/2'. Pass and fail are not equally likely outcomes; the chance depends on preparation. Similarly, 'either it rains or it does not' does not make the probability of rain 1/2.

Working procedure for every problem: (1) describe the experiment and write the sample space, counting n(S); (2) read the event carefully and list the favourable outcomes, counting n(E); (3) divide and simplify; (4) check that the answer is between 0 and 1 and that it makes sense. For instance, if a bag holds 3 red and 5 black balls, the probability of a red ball is 3/8, and it should be less than 1/2 because there are fewer red balls than black.

Probability tells us what to expect over many trials, not what will happen in one. If P(E) = 1/6 for a die showing 4, then in 600 throws we expect about 100 fours, though the actual number may be 94 or 107. The expected number of occurrences in N trials is N × P(E); this simple multiplication is frequently used in problems.

Before the classical definition, people estimated chance by doing the experiment many times. If a trial is repeated n times and an event E happens m times, the experimental probability or empirical probability of E is m/n, the number of times E occurred divided by the number of trials. If a drawing pin is dropped 200 times and lands point-up 84 times, its experimental probability of landing point-up is 84/200 = 0.42. Here the two outcomes, point-up and point-down, are not equally likely, so the classical definition cannot be applied and experiment is the only way.

Experimental probability changes from one set of trials to another. If you toss a coin 10 times you may get 7 heads, giving 0.7; another 10 tosses may give 4 heads, giving 0.4. But as the number of trials grows, the experimental probability settles down close to a fixed value. Experiments recorded by the French naturalist Buffon (4040 tosses, 2048 heads, ratio 0.507), the statistician Karl Pearson (24000 tosses, 12012 heads, ratio 0.5005) and the mathematician J. E. Kerrich, who tossed a coin 10000 times while interned during the Second World War and got 5067 heads (0.5067), all show the ratio approaching 0.5. This tendency is the reason the theoretical value 1/2 is meaningful: it is the value the experimental probability approaches as the number of trials increases indefinitely.

The relation between the two ideas may be summed up thus. Theoretical probability is computed by reasoning, assuming equally likely outcomes; it does not require any experiment. Experimental probability is measured by actually doing the experiment; it requires no assumption about equal likelihood, but it needs a large number of trials to be reliable. Where both can be found, the experimental value approaches the theoretical one for large n.

Experimental probability is the tool of the insurance company, which sets premiums from records of past claims; of the meteorologist, who compares today's conditions with thousands of past days; of the factory, which tests a sample of bulbs to estimate the fraction defective; and of the doctor, who quotes the success rate of a treatment from earlier patients. In examination questions, a table of observed frequencies is given and the student divides the frequency of the required event by the total number of trials.

Two special events mark the ends of the probability scale. An event that is certain to occur is called a sure event or certain event. Its favourable outcomes are all the outcomes, so n(E) = n(S) and P(E) = 1. Getting a number less than 7 when a die is thrown is a sure event: every one of 1, 2, 3, 4, 5, 6 is less than 7. Drawing a ball from a bag that contains only red balls and getting a red ball is a sure event.

An event that cannot possibly occur is called an impossible event. It has no favourable outcome, n(E) = 0, and P(E) = 0. Getting the number 8 on a die, drawing a blue ball from a bag containing only red and white balls, or getting two heads when a single coin is tossed once, are impossible events.

Every other event lies strictly between these two: 0 < P(E) < 1. The nearer the probability is to 1, the more likely the event; the nearer to 0, the less likely. A probability of 1/2 means the event is as likely to happen as not. Thus statements such as 'the probability of rain is 1.2' or 'the probability of winning is −0.3' are meaningless. Examination questions often ask 'which of the following cannot be the probability of an event?' with choices such as 2/3, −1.5, 15 per cent, 0.7; the answer is −1.5 because probability is never negative, and any number greater than 1, such as 1.01 or 105 per cent, is equally impossible.

The sum of the probabilities of all the elementary events of an experiment is 1. For a die, P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 6 × 1/6 = 1. For two coins, P(HH) + P(HT) + P(TH) + P(TT) = 4 × 1/4 = 1. More generally, if the sample space is split into events that do not overlap and together cover every outcome, their probabilities add up to 1. This gives a way to find one unknown probability when the others are known.

A worked check: a bag has some red balls and some blue balls; if P(red) = 0.35, then P(blue) must be 0.65 because red and blue exhaust all cases. If a third colour were present, P(red) + P(blue) would be less than 1, and the remainder would be P(third colour).

Students sometimes confuse an impossible event with an event of very small probability. Winning a lottery with ten lakh tickets has probability 1/1000000, which is tiny but not zero; somebody does win. Drawing a 7 from a die is truly impossible. Likewise a sure event is different from a very likely one: the sun rising tomorrow is treated as certain in everyday speech, but in this chapter P(E) = 1 means that every single outcome of the sample space is favourable. Another common question asks for the probability of an event 'which is certain to happen' or of 'an event that can never happen' when a ball is drawn from a bag with only green balls: P(green) = 1 and P(red) = 0. Whenever the answer to a problem comes out as 0 or 1, the student should be able to say in words why the event is impossible or certain; and if the answer is negative or larger than 1, an arithmetic error has certainly been made and the working must be checked.

For any event E, the event 'E does not occur' is called the complement of E, written as Ē or E′ or 'not E'. If E is 'getting an even number' on a die, then Ē is 'getting an odd number'. If E is 'drawing a red ball', Ē is 'drawing a ball which is not red'. The outcomes favourable to Ē are exactly the outcomes of S that are not in E, so n(Ē) = n(S) − n(E).

Dividing by n(S) gives P(Ē) = 1 − n(E)/n(S) = 1 − P(E). Hence

P(E) + P(Ē) = 1, or P(not E) = 1 − P(E).

Two events E and Ē are called complementary events. Exactly one of them occurs in every trial. The pair is a special case of the rule that probabilities of events that together cover the whole sample space without overlapping add to 1.

This rule saves a great deal of counting. Suppose two dice are thrown and we want the probability that the two numbers are different. Counting pairs with different numbers directly means listing 30 pairs; but the complement, 'both numbers the same', has only 6 outcomes (1,1), (2,2), …, (6,6), so P(same) = 6/36 = 1/6 and P(different) = 1 − 1/6 = 5/6. Whenever the phrase 'at least one' appears, the complement 'none' is usually easier: for three coins, P(at least one head) = 1 − P(no head) = 1 − P(TTT) = 1 − 1/8 = 7/8.

Typical examination questions. (i) The probability that it will rain tomorrow is 0.85; the probability that it will not rain is 1 − 0.85 = 0.15. (ii) If P(E) = 0.05, then P(not E) = 0.95. (iii) A bag has lemon-flavoured candies only; the probability of drawing an orange candy is 0 and of a lemon candy is 1. (iv) Two students Sangeeta and Reshma play a game; the probability of Sangeeta winning is 0.62; since one of them must win and there is no draw, the probability of Reshma winning is 0.38.

Care is needed to describe the complement exactly. The complement of 'at least two heads' in three coins is 'at most one head', not 'no head'. The complement of 'a number greater than 4' on a die is 'a number less than or equal to 4', which includes 4. Writing both sets out in full avoids errors.

Coin problems are the simplest and are used to fix the method. A fair coin has two equally likely outcomes, H and T. For one coin, P(H) = P(T) = 1/2.

Two coins tossed together (or one coin tossed twice) give the sample space {HH, HT, TH, TT} with n(S) = 4. Then P(two heads) = P(HH) = 1/4; P(exactly one head) = P({HT, TH}) = 2/4 = 1/2; P(at least one head) = P({HH, HT, TH}) = 3/4; P(no head) = P(TT) = 1/4; P(at most one head) = P({HT, TH, TT}) = 3/4. The typical mistake is to treat 'one head and one tail' as a single outcome and answer 1/3; the two coins are distinct and HT and TH are different outcomes.

Three coins tossed together give 8 outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Then P(three heads) = 1/8; P(exactly two heads) = P({HHT, HTH, THH}) = 3/8; P(exactly one head) = 3/8; P(no head) = 1/8; P(at least two heads) = 4/8 = 1/2; P(at least one head) = 7/8; P(all the same face) = P({HHH, TTT}) = 2/8 = 1/4.

A well-known problem: Hanif wins a game if all the tosses of a coin give the same result, that is three heads or three tails, and loses otherwise. The probability that he loses is 1 − 2/8 = 6/8 = 3/4.

Why the toss is fair. When a cricket match starts, the captains toss a coin because head and tail are equally likely, so neither team has an advantage. If instead a die were thrown and one captain given the numbers 1 and 2 and the other 3, 4, 5, 6, the second captain would win with probability 4/6, which is unfair. Also note that a coin's past tosses do not influence its next toss; ten heads in a row do not make a tail 'due'.

A coin tossed n times has 2ⁿ outcomes. Writing them in a systematic order, for instance by treating H as 0 and T as 1 and counting in binary, ensures that none is missed. For problems asking for 'the same result on all tosses' there are always exactly two favourable outcomes, so the probability is 2/2ⁿ = 1/2ⁿ⁻¹.

A die is a cube with faces numbered 1 to 6. A fair or unbiased die has all six outcomes equally likely, so each has probability 1/6.

One die. Events are described by properties of the number: P(even) = 3/6 = 1/2; P(prime) = P({2, 3, 5}) = 1/2; P(multiple of 3) = P({3, 6}) = 1/3; P(number greater than 4) = P({5, 6}) = 1/3; P(number between 2 and 6) = P({3, 4, 5}) = 1/2; P(perfect square) = P({1, 4}) = 1/3; P(a number less than 1) = 0.

Two dice thrown together, or one die thrown twice, give 36 equally likely ordered pairs (a, b), where a is the number on the first die and b on the second. The best way to handle these problems is to draw a 6 × 6 table. The most common question concerns the sum of the two numbers. The possible sums are 2 to 12, and their favourable counts are: sum 2: 1, sum 3: 2, sum 4: 3, sum 5: 4, sum 6: 5, sum 7: 6, sum 8: 5, sum 9: 4, sum 10: 3, sum 11: 2, sum 12: 1. These add to 36. So P(sum 7) = 6/36 = 1/6 is the largest, and P(sum 2) = P(sum 12) = 1/36 the smallest. A student who says 'there are 11 possible sums, so each has probability 1/11' has made the classic error: the sums are not equally likely.

Other two-dice events: P(doublet, both numbers equal) = 6/36 = 1/6; P(sum is at least 10) = P({(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)}) = 6/36 = 1/6; P(sum ≤ 4) = (1 + 2 + 3)/36 = 6/36 = 1/6; P(5 does not appear on either die) = 5 × 5/36 = 25/36, so P(5 appears at least once) = 11/36; P(5 appears on exactly one die) = 10/36 = 5/18; P(product is 12) = P({(2,6), (3,4), (4,3), (6,2)}) = 4/36 = 1/9; P(the first number is greater than the second) = 15/36 = 5/12.

A die may also carry letters or symbols instead of numbers. If the faces show A, B, C, D, E and A, then P(A) = 2/6 = 1/3 and P(D) = 1/6. Whatever is written, the six faces remain the six equally likely outcomes.

Finally, an odd-shaped object such as a matchbox or a drawing pin thrown on the floor does not have equally likely outcomes; its probabilities must be found experimentally.

A standard pack, or deck, of playing cards has 52 cards. They are divided into four suits of 13 cards each: spades (♠) and clubs (♣), which are black, and hearts (♥) and diamonds (♦), which are red. So there are 26 black cards and 26 red cards. Each suit has the cards ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen and king. The jack, queen and king are called face cards (or court cards or picture cards); there are 3 per suit, hence 12 in all. Sometimes the ace is grouped with them as an 'honour card', but in this chapter face cards means only jack, queen and king. There are 4 aces, 4 kings, 4 queens and so on, one of each suit. A pack normally has no jokers unless the problem says so.

When a card is drawn at random from a well-shuffled pack, n(S) = 52 and the outcomes are equally likely. Then P(a king) = 4/52 = 1/13; P(a red card) = 26/52 = 1/2; P(a spade) = 13/52 = 1/4; P(a face card) = 12/52 = 3/13; P(a red face card) = 6/52 = 3/26; P(the king of hearts) = 1/52; P(a black king) = 2/52 = 1/26; P(a jack of clubs or a queen of diamonds) = 2/52 = 1/26; P(neither a king nor a queen) = 1 − 8/52 = 44/52 = 11/13; P(a card that is a 10 or a spade) = (13 + 3)/52 = 16/52 = 4/13, because the 10 of spades is already counted among the spades and must not be counted twice.

Problems sometimes remove cards before drawing. If the king, queen and jack of clubs are removed and then a card is drawn, n(S) = 49; P(a heart) = 13/49; P(a king) = 3/49; P(a club) = 10/49. If the five cards ten, jack, queen, king and ace of diamonds are set aside and one card is drawn from the remaining 47, P(a queen) = 3/47. If the queen is then set aside too and a second card drawn from 46, P(an ace) = 3/46 and P(a queen) = 2/46 = 1/23. Always recompute the total after every removal.

Small sets of cards are also used. Five cards, the ten, jack, queen, king and ace of diamonds, are shuffled face down and one is picked; P(a queen) = 1/5. If the queen is put aside and another card picked, P(an ace) = 1/4 and P(a queen) = 0. Or cards numbered 2 to 101 are placed in a box: n(S) = 100, P(an even number) = 50/100 = 1/2, P(a number which is a perfect square) = P({4, 9, 16, 25, 36, 49, 64, 81, 100}) = 9/100, P(a prime number less than 20) = P({2, 3, 5, 7, 11, 13, 17, 19}) = 8/100 = 2/25.

A large family of problems concerns drawing one object from a collection: balls or marbles from a bag, bulbs or pens from a box, tickets or discs numbered in a lottery, or choosing one person from a group. Because each object is equally likely to be picked, n(S) is the total number of objects and n(E) is the number having the required property.

Bags of balls. A bag has 3 red and 5 black balls; a ball is drawn at random. P(red) = 3/8, P(not red) = 5/8. A box has 5 red, 8 white and 4 green marbles; P(red) = 5/17, P(white) = 8/17, P(not green) = 13/17.

Finding an unknown number. A bag contains 5 red balls and some blue balls; the probability of drawing a blue ball is double that of a red ball. Let the number of blue balls be x. Then x/(5 + x) = 2 × 5/(5 + x), so x = 10. Again, a box holds 12 balls of which x are black; if 6 more black balls are added, the probability of a black ball doubles: (x + 6)/18 = 2 × x/12, giving 12x + 72 = 36x, 24x = 72, x = 3. Such problems turn the definition into an equation.

Lots and defective items. A lot of 20 bulbs has 4 defective ones. P(a bulb drawn is defective) = 4/20 = 1/5. If the drawn bulb is not defective and is not replaced, P(the next bulb is not defective) = 15/19: both the total and the favourable count fall by one. A lot has 144 ball pens of which 20 are defective; Nuri buys a pen only if it is good. P(she buys) = 124/144 = 31/36; P(she does not buy) = 20/144 = 5/36.

Choosing a person. A group of 3 students has 2 girls and 1 boy; if the class representative is chosen by lot, P(a girl) = 2/3. In a school of 40 students, 20 are in Class 10, 12 in Class 9 and 8 in Class 8; P(the student chosen is not from Class 10) = 20/40 = 1/2.

Numbered tickets. A piggy bank has 100 fifty-paise coins, 50 one-rupee coins, 20 two-rupee coins and 10 five-rupee coins; when turned upside down, one coin falls out. P(fifty-paise coin) = 100/180 = 5/9; P(not a five-rupee coin) = 170/180 = 17/18. A box has cards numbered 1 to 20; P(a multiple of 4) = 5/20 = 1/4. Ninety discs numbered 1 to 90: P(two-digit number) = 81/90 = 9/10; P(perfect square) = 9/90 = 1/10; P(divisible by 5) = 18/90 = 1/5.

In every case, list or count carefully, watch for the word 'not', and check whether the total has changed because of a removal.

So far every sample space was a finite set of outcomes that we could count. Sometimes the outcomes are all the points of a line segment, a region of a plane or a solid, and cannot be counted; then we measure instead. If a point is chosen at random in a region of area A, and E is the event that it falls in a part of area a, the geometric probability is P(E) = a/A, the ratio of the favourable area to the total area. Similarly for lengths: if a point is chosen at random on a segment of length L, the probability that it lies in a piece of length l is l/L. The assumption is that every point of the region is equally likely, so equal areas have equal chances.

Musical chairs. In a game of musical chairs the person in charge stops the music at any moment within 2 minutes after starting. What is the probability that the music stops within the first half-minute? Every instant in the 2 minutes (120 seconds) is equally likely. The favourable interval is the first 30 seconds, so P = 30/120 = 1/4.

The dropped helicopter. A missing helicopter is reported to have crashed somewhere in a rectangular region 4.5 km by 9 km, inside which lies a lake 3 km by 2.5 km. The probability that it crashed in the lake is (area of lake)/(area of region) = (3 × 2.5)/(4.5 × 9) = 7.5/40.5 = 5/27.

Circle in a rectangle. A rectangular lawn 3 m by 2 m has a circular flower bed of diameter 1 m in it. A ball is thrown and lands somewhere on the lawn. P(it lands in the flower bed) = π(0.5)² / (3 × 2) = (22/7 × 0.25)/6 = 0.7857/6 ≈ 0.131. A dart thrown at a square board of side 20 cm with a circular target of radius 5 cm has P(hit) = π × 25/400 = 25 × 22/(7 × 400) ≈ 0.196.

Points on a segment. A point is chosen at random on a scale from 0 to 10; P(it lies between 3 and 5) = 2/10 = 1/5.

Geometric probability shares all the properties of ordinary probability: it is between 0 and 1, the whole region has probability 1, and P(not E) = 1 − P(E). Where units are involved, both areas must be in the same unit so that they cancel. Any shape may appear: a triangle inside a rectangle, a smaller square inside a larger, a circular ring, or a shaded part of a figure; the student only needs the area formulas of Class 9 and 10.

The Telangana SSC examination asks probability questions of one, two and four marks. One-mark items test definitions and the range 0 to 1, for example 'the probability of a sure event is ___' or 'if P(E) = 0.3 then P(not E) = ___'. Two-mark items ask a single computation, such as the probability of a face card. Four-mark items combine two or three parts, often with two dice, three coins or a bag whose numbers must be found. The following solutions illustrate the complete method.

Problem 1. A bag contains 5 red, 4 green and 3 black balls; one is drawn at random. Find the probability that it is (i) green, (ii) not red, (iii) red or black. Solution: n(S) = 12. (i) P(green) = 4/12 = 1/3. (ii) P(not red) = 7/12. (iii) P(red or black) = 8/12 = 2/3. Check: P(green) + P(red or black) = 1/3 + 2/3 = 1.

Problem 2. Two dice are thrown at the same time. Find the probability that the sum is (i) 8, (ii) 13, (iii) less than or equal to 12. Solution: n(S) = 36. (i) Favourable: (2,6), (3,5), (4,4), (5,3), (6,2), so 5/36. (ii) No outcome, so 0. (iii) Every outcome, so 1.

Problem 3. Gopi buys a fish from a shop for his aquarium. The tank has 5 male and 8 female fish and the shopkeeper takes one out at random. P(male fish) = 5/13.

Problem 4. A game of chance consists of spinning an arrow that comes to rest pointing at one of the numbers 1 to 8, all equally likely. P(8) = 1/8; P(odd number) = 4/8 = 1/2; P(number greater than 2) = 6/8 = 3/4; P(number less than 9) = 1.

Problem 5. A child has a die whose six faces show the letters A, B, C, D, E, A. P(A) = 2/6 = 1/3; P(D) = 1/6.

Problem 6. Suppose you drop a die at random on the rectangular region 3 m by 2 m shown in a figure which has a circle of diameter 1 m inside. P(the die lands inside the circle) = π × 0.25/6 = π/24.

Problem 7. A bag has 5 red balls and some blue balls. If P(blue) is double P(red), find the number of blue balls: x/(5 + x) = 10/(5 + x), so x = 10.

Method. (1) Identify the experiment; state n(S). (2) List favourable outcomes; state n(E). (3) Write P(E) = n(E)/n(S) and simplify. (4) For 'not', 'at least', 'neither…nor', consider the complement. (5) Write the answer as a fraction in lowest terms and check 0 ≤ P ≤ 1. Marks are given for the sample space and the listing of favourable outcomes, so show them, even in a two-mark question.