Chapter 14 — Probability

Every day we make statements about uncertainty. A farmer says 'it will probably rain this week', a cricket commentator says 'India has a good chance of winning', a doctor says 'this medicine usually works'. Each statement carries an idea of likelihood, but the words probably, good chance and usually are not precise. Two people can hear the same words and understand very different amounts of confidence. Mathematics removes this vagueness by assigning a number to the likelihood of an event. That number is called the probability of the event.

The probability of any event is always a number from 0 to 1. A probability of 0 means the event cannot happen at all; a probability of 1 means the event is certain to happen. Values in between show degrees of chance: a probability of 0.5 means the event is as likely to happen as not, and a probability of 0.9 means it is very likely. Probability can be written as a fraction, a decimal or a percentage; 1/4, 0.25 and 25% mean exactly the same thing.

The subject grew out of questions about gambling. In the seventeenth century the French mathematicians Blaise Pascal and Pierre de Fermat exchanged letters about how to divide the stakes of an unfinished dice game fairly. Their reasoning became the foundation of the theory of probability. Later the Italian mathematician Girolamo Cardano, the Swiss family of the Bernoullis and the French mathematician Pierre-Simon Laplace developed the subject into a complete branch of mathematics.

Today probability is far more than a tool for games. Insurance companies fix premiums using the probability of accidents. Weather departments forecast '70% chance of rain'. Doctors compare the probability of recovery under different treatments. Quality-control engineers test a small sample of bulbs to estimate the chance that a bulb from the factory is defective. Genetics predicts the chance that a child inherits a particular trait. In all these cases the same simple rules that you will learn with coins and dice are used.

In this chapter you will first see how probability is measured by experiment, then how it is calculated by pure reasoning when the outcomes are equally likely, and finally how the two views agree when an experiment is repeated a very large number of times.

One way to measure chance is simply to try the experiment many times and count. If a coin is tossed 100 times and a head appears 46 times, we say the experimental probability of a head in this set of trials is 46/100 = 0.46. This is also called the empirical probability, because it is based on observation (empirical means 'based on experience').

The rule is: experimental probability of an event E = (number of trials in which E happened) ÷ (total number of trials). The number of trials must be counted honestly and every trial must be performed under the same conditions.

Consider a class activity. Each of 40 students tosses a coin 10 times, giving 400 tosses in all. Suppose heads appear 208 times. The experimental probability of a head is 208/400 = 0.52. If the same class tosses again, they may get 195 heads, giving 0.4875. The experimental probability changes from one set of trials to another; it is not a fixed number. But notice that both values are close to 1/2, and as the number of tosses grows the value comes closer and closer to 1/2. The English statistician Karl Pearson tossed a coin 24,000 times and got 12,012 heads, a proportion of 0.5005.

The same idea applies to a die. If a die is thrown 600 times and the number 4 appears 97 times, the experimental probability of getting a 4 is 97/600 ≈ 0.162, which is close to 1/6 ≈ 0.167.

Experimental probability is the only method available when we cannot reason about equally likely outcomes. A factory does not know beforehand how many bulbs will be defective; it tests 1000 bulbs, finds 12 defective, and estimates the probability of a defective bulb as 12/1000 = 0.012. A cricket board estimates a batsman's chance of scoring fifty from his past innings. A weather office estimates the probability of rain on a given date from decades of records.

The limitation of the experimental method is that it needs many trials and still gives only an estimate. This is why, whenever the outcomes can be assumed equally likely, we prefer the theoretical method described next.

To calculate probability by reasoning, we need precise words. An experiment is any action whose result is observed. It is called a random experiment if it has more than one possible result and we cannot predict with certainty which result will occur, even though we know all the possible results in advance. Tossing a coin, throwing a die, drawing a card from a shuffled pack and picking a ball from a bag without looking are all random experiments. Heating water to 100°C at sea level is not a random experiment, because the result (boiling) is certain.

Each possible result of a random experiment is called an outcome. The set of all possible outcomes is called the sample space, usually denoted S. Writing the sample space correctly is the first step in every probability problem.

• One coin: S = {H, T}, 2 outcomes.
• One die: S = {1, 2, 3, 4, 5, 6}, 6 outcomes.
• Two coins: S = {HH, HT, TH, TT}, 4 outcomes. Note that HT and TH are different outcomes.
• Three coins: S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, 8 outcomes.
• Two dice: 36 outcomes, written as ordered pairs (1,1), (1,2), …, (6,6).

A common mistake is to say that two coins have only three outcomes: 'two heads, two tails, one of each'. These three are not equally likely, because 'one of each' can happen in two ways (HT and TH). When we use the classical formula the outcomes must be equally likely, so the correct sample space has four members.

A useful pattern: if an experiment is made of k stages and each stage has n equally likely results, the sample space has n^k outcomes. Two coins give 2² = 4, three coins give 2³ = 8, two dice give 6² = 36, and a coin with a die gives 2 × 6 = 12.

Outcomes are called equally likely when there is no reason to expect any one of them more than the others. A fair coin, an unbiased die, a well-shuffled pack and balls that are identical except in colour all give equally likely outcomes. A biased die or a coin with a heavier side does not, and then the classical formula cannot be used.

An event is any collection of outcomes of a random experiment, that is, any 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 prime number' is G = {2, 3, 5}. We say the event has occurred if the outcome of the experiment belongs to the event.

An event that contains exactly one outcome is called an elementary event (or simple event). For a die, the elementary events are {1}, {2}, {3}, {4}, {5} and {6}. For two coins, they are {HH}, {HT}, {TH} and {TT}. Every other event is a union of elementary events; for instance, the event 'at least one head' with two coins is {HH, HT, TH}, made of three elementary events.

An important property: the sum of the probabilities of all elementary events of an experiment is 1. For a fair die, each elementary event has probability 1/6, and 6 × (1/6) = 1. For two coins, 4 × (1/4) = 1. This property is a check on your work: if your elementary probabilities do not add up to 1, something has been listed wrongly.

Two events with no common outcome are called mutually exclusive. 'Getting an even number' and 'getting an odd number' on a die cannot happen together. 'Getting an even number' and 'getting a prime number' are not mutually exclusive, because 2 is both. Events that together cover the whole sample space are called exhaustive. Even and odd are both mutually exclusive and exhaustive for a die.

Consider the event 'getting a number less than 7' on a die. Every outcome satisfies it, so the event equals the whole sample space. This is a sure event (or certain event) and its probability is 1. The event 'getting an 8' has no outcome at all; it is the empty set, an impossible event, with probability 0.

When solving a problem, first write the sample space, then write the event as a set of outcomes, then count. Writing the event as a set prevents the common errors of double counting or missing an outcome. For example, 'a number divisible by 2 or 3' on a die is {2, 3, 4, 6}, four outcomes, not five, because 6 must be counted only once.

When all outcomes of a random experiment are equally likely, the probability of an event can be calculated without doing the experiment at all. This is the theoretical or classical probability, given by

P(E) = (number of outcomes favourable to E) ÷ (total number of equally likely outcomes)

The outcomes that belong to the event are called favourable outcomes. The word favourable does not mean good; it only means 'belonging to the event we are asking about'. If the event is 'getting a defective bulb', the defective bulbs are the favourable outcomes.

Take a fair die. The total number of outcomes is 6. For the event 'getting a 5', the favourable outcomes are just one, so P(5) = 1/6. For 'getting an even number', the favourable outcomes are 2, 4, 6, so P(even) = 3/6 = 1/2. For 'getting a number less than 3', the outcomes are 1 and 2, so the probability is 2/6 = 1/3. Always reduce the fraction to lowest terms unless the question asks for a decimal or percentage.

The formula rests on two conditions. First, the outcomes must be equally likely; second, the outcomes must be mutually exclusive and exhaustive, that is, exactly one of them happens on every trial. If a spinner has three sectors of unequal sizes, the three outcomes are not equally likely and the formula with 'total = 3' is wrong; we would instead compare angles or areas, as shown later in the chapter.

Because the number of favourable outcomes is between 0 and the total number of outcomes, the probability always lies between 0 and 1. If the favourable outcomes are all the outcomes, the probability is 1 (sure event). If there is no favourable outcome, the probability is 0 (impossible event).

The theoretical and experimental probabilities are connected by a principle known as the law of large numbers: if a random experiment is repeated a very large number of times, the experimental probability of an event comes closer and closer to its theoretical probability. This is why a coin tossed 24,000 times gave almost exactly half heads. The theoretical value is what the experiment 'tends towards'.

A useful habit for examinations is to write three lines every time: 'Total outcomes = …', 'Favourable outcomes = …', 'P(E) = … / … = …'. This shows the examiner the reasoning and earns method marks even if a slip is made in counting.

For every event E there is the event 'E does not happen', written as not E, E' or Ē. It is called the complementary event (or complement) of E. If E is 'getting an even number' on a die, then Ē is 'getting an odd number'. If E is 'drawing a king' from a pack, Ē is 'drawing a card that is not a king'.

Since E and Ē together contain every outcome and have no outcome in common, the number of favourable outcomes of Ē is (total outcomes) − (favourable outcomes of E). Dividing by the total gives

P(Ē) = 1 − P(E), or equivalently P(E) + P(Ē) = 1.

This rule saves an enormous amount 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. Instead, count the pairs with the same number: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), six pairs. P(same) = 6/36 = 1/6, so P(different) = 1 − 1/6 = 5/6.

Again, if a bag has 3 red balls and 5 black balls, P(red) = 3/8, so P(not red) = 1 − 3/8 = 5/8, which agrees with the direct count 5/8 for black. The rule is especially useful when the event is described with 'at least' or 'not'. 'At least one head' in three tosses is the complement of 'no head', i.e. TTT. P(no head) = 1/8, so P(at least one head) = 7/8.

The rule also gives a check: probabilities of E and its complement must add to exactly 1. If you compute P(E) = 0.3 and P(Ē) = 0.6, one of them is wrong.

Do not confuse 'complementary' with 'mutually exclusive'. Complementary events are always mutually exclusive, but mutually exclusive events need not be complementary. 'Getting a 1' and 'getting a 2' on a die are mutually exclusive, but they are not complements of each other because their probabilities add to 2/6, not 1.

A typical examination question gives P(E) and asks for P(not E), or gives P(not E) = 0.35 and asks for P(E) = 0.65. Another common form: 'The probability that it will rain tomorrow is 0.85; what is the probability that it will not rain?' The answer is 1 − 0.85 = 0.15.

Coin problems train the habit of listing the sample space fully. A fair coin has two equally likely faces, head (H) and tail (T).

One coin. S = {H, T}. P(H) = 1/2, P(T) = 1/2.

Two coins tossed together (or one coin tossed twice). S = {HH, HT, TH, TT}, four equally likely outcomes.

• P(two heads) = 1/4.
• P(exactly one head) = 2/4 = 1/2, from HT and TH.
• P(at least one head) = 3/4, from HH, HT, TH; or 1 − P(TT) = 1 − 1/4 = 3/4.
• P(at most one head) = 3/4, from HT, TH, TT.
• P(no head) = 1/4.

Notice that 'at least one' means one or more, while 'at most one' means one or fewer. Students often mix these up. Read the phrase, translate it into a set of outcomes, then count.

Three coins. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}, eight outcomes. A systematic way to list them is by the number of heads: three heads (1 way), two heads (3 ways), one head (3 ways), no head (1 way), giving 1 + 3 + 3 + 1 = 8.

• P(three heads) = 1/8.
• P(exactly two heads) = 3/8.
• P(at least two heads) = 4/8 = 1/2 (three heads or two heads: 1 + 3 = 4 outcomes).
• P(at least one tail) = 1 − P(HHH) = 7/8.
• P(same result on all three) = 2/8 = 1/4 (HHH or TTT).

A question that appears in many forms: 'Hanif wins if all three tosses give the same result, otherwise he loses. What is the probability that Hanif loses?' P(all same) = 2/8 = 1/4, so P(loses) = 1 − 1/4 = 3/4.

Another form uses a coin and a die together. The 12 outcomes are H1 to H6 and T1 to T6. P(head and an even number) = 3/12 = 1/4. P(tail or a number greater than 4) counts T1–T6 (6 outcomes) plus H5, H6 (2 more) = 8/12 = 2/3.

The key skill is complete and non-repeating listing. When the number of coins is n, the sample space has 2ⁿ outcomes; check that your list has that many before counting favourable ones.

A die is a cube with faces numbered 1 to 6. An unbiased die gives six equally likely outcomes.

One die. Typical events and their probabilities: P(even) = 3/6 = 1/2; P(prime) = 3/6 = 1/2 (2, 3, 5); P(multiple of 3) = 2/6 = 1/3; P(number less than 5) = 4/6 = 2/3; P(6) = 1/6; P(number greater than 6) = 0; P(number less than 7) = 1.

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. Draw a 6 × 6 table and write the sum a + b in each cell; this table answers most questions.

The number of pairs giving each sum is: 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. Check: 1+2+3+4+5+6+5+4+3+2+1 = 36. Seven is the most likely sum, with probability 6/36 = 1/6.

• P(sum 8) = 5/36, from (2,6), (3,5), (4,4), (5,3), (6,2).
• P(sum ≤ 4) = (1 + 2 + 3)/36 = 6/36 = 1/6.
• P(sum 13) = 0; P(sum ≤ 12) = 1.
• P(doublet, i.e. same number on both) = 6/36 = 1/6.
• P(5 on at least one die) = 11/36, counting (5,1)…(5,6) and (1,5)…(6,5) with (5,5) once: 6 + 6 − 1 = 11.
• P(product 12) = 4/36 = 1/9, from (2,6), (3,4), (4,3), (6,2).
• P(sum is a prime) = pairs with sum 2, 3, 5, 7, 11 = 1 + 2 + 4 + 6 + 2 = 15, so 15/36 = 5/12.

The common error is to treat (2,6) and (6,2) as the same outcome, giving 21 outcomes. They are different results of the experiment, and only the 36 ordered pairs are equally likely. If a question says 'two dice, one blue and one grey', it is reminding you that order matters.

Another type of question: 'A die is thrown twice. Find the probability that 5 will not come up either time.' P(no 5 on one throw) = 5/6, and counting pairs with neither entry 5 gives 5 × 5 = 25 pairs, so P = 25/36. Then P(5 comes up at least once) = 1 − 25/36 = 11/36.

A standard deck has 52 cards in four suits: spades (♠) and clubs (♣), which are black, and hearts (♥) and diamonds (♦), which are red. Each suit has 13 cards: ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen and king. So there are 26 red cards and 26 black cards, 4 cards of each denomination (four aces, four kings and so on) and 13 cards of each suit.

The jack, queen and king are called face cards (or picture cards or court cards). There are 3 face cards per suit, so 12 face cards in all. In many board problems the ace is not counted as a face card. The cards 2 to 10 are number cards, 9 per suit, 36 in all.

With a well-shuffled deck, drawing one card gives 52 equally likely outcomes. Some standard results:

• P(an ace) = 4/52 = 1/13.
• P(a red card) = 26/52 = 1/2.
• P(a heart) = 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 king or a queen) = 8/52 = 2/13.
• P(a spade or an ace) = 13 + 4 − 1 = 16 outcomes, so 16/52 = 4/13. The ace of spades is counted once only.
• P(neither a jack nor a king) = 1 − 8/52 = 44/52 = 11/13.
• P(a black card or a queen) = 26 + 4 − 2 = 28, so 28/52 = 7/13, since two queens are black.

Sometimes cards are removed before the draw, which changes the total. If the king, queen and jack of clubs are removed, 49 cards remain. Then P(a heart) = 13/49, P(a queen) = 3/49, P(a club) = 10/49, P(the ten of hearts) = 1/49. If all the face cards are removed, 40 cards remain and P(an ace) = 4/40 = 1/10, P(a black number card) = 18/40 = 9/20.

Read the wording carefully. 'A red king' means one of two cards; 'a red card or a king' means 26 + 2 = 28 cards; 'a card that is red and a king' means 2 cards. The words 'or' and 'and' change the count, and cards that satisfy both conditions must not be counted twice when 'or' is used.

Many examination questions describe a bag or box containing objects that differ only in colour or number. Since the objects are identical in size and shape and the draw is made without looking, each object is equally likely to be picked. The total number of outcomes equals the number of objects.

Balls of different colours. A bag contains 3 red, 5 black and 4 white balls, 12 in all. P(red) = 3/12 = 1/4; P(black) = 5/12; P(not white) = 8/12 = 2/3; P(red or white) = 7/12; P(blue) = 0.

Finding an unknown number. A bag has 5 red balls and x black balls. If the probability of drawing a black ball is double that of a red ball, find x. P(red) = 5/(5 + x), P(black) = x/(5 + x). Given x/(5 + x) = 2 × 5/(5 + x), so x = 10. Questions of this kind lead to a simple equation and are frequent in the public examination.

Another version: a jar has 24 marbles, some green and the rest blue. If the probability of a green marble is 2/3, the number of green marbles is 24 × 2/3 = 16, so 8 are blue.

Lots and defective items. A lot of 20 bulbs contains 4 defective ones. One bulb is drawn at random. P(defective) = 4/20 = 1/5. If the drawn bulb is not defective and is not replaced, 19 bulbs remain with 4 defective, so P(next bulb not defective) = 15/19. Notice how the total and the favourable count both change after removal without replacement.

Numbered tickets or discs. A box contains cards numbered 3, 4, 5, …, 50, which is 48 cards. P(number divisible by 7) counts 7, 14, 21, 28, 35, 42, 49, that is 7 cards, so 7/48. P(a perfect square) counts 4, 9, 16, 25, 36, 49, 6 cards, so 6/48 = 1/8. Count the total carefully: from 3 to 50 inclusive there are 50 − 3 + 1 = 48 numbers.

Pens in a lot. 12 defective pens are mixed with 132 good ones, 144 in all. P(good pen) = 132/144 = 11/12.

Reasoning about fairness. If a buyer of shirts accepts one only when it has no defect, and 88 of 100 shirts are good, the probability that a shirt is acceptable to her is 88/100 = 0.88. If another buyer accepts shirts with minor defects too (8 more), his probability is 96/100 = 0.96.

Calendar problems test counting more than formula. Two facts are needed: a non-leap year has 365 days = 52 weeks + 1 day, and a leap year has 366 days = 52 weeks + 2 days.

Sundays in a non-leap year. The 52 full weeks contain 52 Sundays for certain. The one extra day can be any of the seven days of the week, equally likely. If it is a Sunday, the year has 53 Sundays. So P(53 Sundays) = 1/7 and P(52 Sundays) = 6/7.

Sundays in a leap year. The two extra days are consecutive: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat) or (Sat, Sun), seven equally likely pairs. Two of them contain a Sunday, so P(53 Sundays) = 2/7. Similarly P(53 Mondays) = 2/7 and P(53 Sundays and 53 Mondays) = 1/7 (only the pair Sun, Mon).

Days of a month. If a date in June (30 days) is chosen at random, P(the date is a multiple of 5) = 6/30 = 1/5, from 5, 10, 15, 20, 25, 30. February in a leap year has 29 days, so P(a Sunday) from a February that starts on a Sunday is 5/29, because Sundays fall on 1, 8, 15, 22 and 29.

Birthdays. Two friends were born in the same non-leap year. What is the probability that their birthdays are (i) the same day, (ii) different days? For the first friend the birthday can be any of 365 days; the second friend has 365 equally likely days, of which only 1 matches. So P(same) = 1/365 and P(different) = 1 − 1/365 = 364/365.

Weekdays. A day of the week is chosen at random: P(it starts with the letter T) = 2/7 (Tuesday, Thursday); P(it is a weekend day) = 2/7; P(it is not Sunday) = 6/7.

Months. If a month of the year is chosen, P(it has 31 days) = 7/12; P(it begins with J) = 3/12 = 1/4.

The technique in all these problems is the same: identify the equally likely outcomes (7 days, 12 months, 30 dates, 365 birthdays), write the favourable outcomes explicitly, and divide. When a problem says 'chosen at random' it is telling you that the outcomes are equally likely.

So far the outcomes have been countable: faces of a die, cards, balls. Some experiments have infinitely many outcomes that cannot be counted, such as the exact point where a dart lands on a board or the exact time a bus arrives. For these, the ratio of counts is replaced by a ratio of measures, that is, of lengths, areas or volumes. This is called geometric probability.

If a point is chosen at random inside a region of area A, the probability that it lies inside a smaller region of area a within it is a/A. 'At random' here means every point is equally likely, so likelihood is proportional to area.

Dartboard. A circular board of radius 12 cm has a bull's-eye of radius 3 cm at its centre. A dart thrown at random hits the board. P(bull's-eye) = (area of small circle)/(area of large circle) = π(3)²/π(12)² = 9/144 = 1/16. The π cancels; only the squares of the radii matter.

Circle in a square. A circle of diameter 1 m is drawn inside a square of side 3 m. A die is dropped on the square. P(it lands inside the circle) = π(0.5)²/(3 × 3) = (π/4)/9 = π/36 ≈ 0.087.

Rectangle in a rectangle. A rectangular garden 20 m by 16 m has a rectangular pond 8 m by 5 m. A ball thrown at random lands in the garden. P(in the pond) = (8 × 5)/(20 × 16) = 40/320 = 1/8.

Spinner with unequal sectors. A wheel is divided into sectors of 90°, 120° and 150°. It is spun once. P(pointer stops in the 120° sector) = 120/360 = 1/3. Angles are used because the area of a sector is proportional to its angle. When sectors are equal, say 8 equal sectors numbered 1 to 8, the counting method works again: P(number greater than 6) = 2/8 = 1/4.

Length. A point is marked at random on a line segment of length 10 cm. P(it lies within 2 cm of one particular end) = 2/10 = 1/5. A bus arrives at a stop at a random moment between 9:00 and 9:30; P(it arrives in the first 10 minutes) = 10/30 = 1/3.

Geometric probability makes it clear that probability 0 does not always mean 'impossible': the chance that the dart lands exactly on the centre point is 0, since a point has no area, yet it is not impossible. Still, for school purposes, we say P = 0 for an impossible event and P = 1 for a sure event.

The classical formula is valid only for equally likely outcomes, so a student must learn to judge when that condition holds. The board asks this in the form 'Which of the following experiments have equally likely outcomes? Explain.'

A driver attempts to start a car. The car starts or does not start. These two outcomes are not equally likely; a well-maintained car starts almost every time. The formula P(starts) = 1/2 would be absurd.

A player attempts to shoot a basketball. She hits or misses. Not equally likely; the chance depends on her skill and distance.

A trial is made to answer a true-false question; the answer is right or wrong. If the person guesses blindly, the two outcomes are equally likely and P(right) = 1/2. If the person knows the subject, they are not.

A baby is born; it is a boy or a girl. These are taken as equally likely in school problems, so P(boy) = 1/2, though real birth ratios are very slightly different.

Tossing a coin is a fair way to decide who bats first. A fair coin gives H and T each probability 1/2, so neither team is favoured. The same reasoning tells us why dice games use unbiased dice and why a 'loaded' die is cheating.

The spinner argument. A spinner has three regions labelled A, B, C but with unequal angles. A student says P(A) = 1/3 'because there are three outcomes'. This is wrong; the correct probabilities are the angles divided by 360°.

The two-coin argument. Someone says that with two coins there are three outcomes, two heads, two tails, one of each, so each has probability 1/3. The error is that 'one of each' occurs in two ways, HT and TH, and is twice as likely as either of the others. The correct equally likely outcomes are HH, HT, TH, TT with probability 1/4 each, and P(one of each) = 1/2.

Sum of two dice. The sums 2 to 12 are eleven outcomes, but they are not equally likely; sum 7 occurs in six ways and sum 2 in only one. The equally likely outcomes are the 36 ordered pairs.

The general test: an outcome set is equally likely when it comes from a fair device (unbiased coin, unbiased die, well-shuffled pack, identical balls, random point) and when each listed outcome is a single indivisible result rather than a group of results of different sizes. Whenever an outcome can be split into sub-outcomes, split it, and then check again.

Probability questions are short, but marks are lost through small errors. Here are the errors seen most often and how to avoid them.

1. Wrong total. Counting 21 outcomes for two dice instead of 36, or 3 outcomes for two coins instead of 4. Always use ordered outcomes and check the count with the rule n^k.

2. Forgetting removed cards or balls. If cards are removed from a deck, the total is no longer 52 and the favourable count may also change. Recount both.

3. Double counting with 'or'. 'A king or a red card' has 2 + 26 = 28 favourable cards, because the two red kings are already among the 26 red cards; adding 4 kings to 26 red cards gives 30, which is wrong. Use n(A) + n(B) − n(A and B).

4. Misreading 'at least' and 'at most'. At least two heads in three tosses means two or three heads (4 outcomes); at most two heads means zero, one or two heads (7 outcomes).

5. Probability outside [0, 1]. If your answer is 7/6 or −1/4 you have made an arithmetic error; probability cannot exceed 1 or be negative.

6. Not simplifying. Give 3/6 as 1/2 and 12/52 as 3/13. A decimal is acceptable if the question asks for it.

7. Ace counted as a face card. Unless the question says so, face cards are only J, Q, K, twelve in all.

8. Leap year confusion. Ordinary year: 1 extra day; leap year: 2 consecutive extra days.

Examination strategy. The Class 10 public examination sets probability in every section: one-mark objective questions such as 'the probability of an impossible event is ____', two-mark questions such as finding P(not E) or a simple card draw, four-mark questions on two dice or bags with an unknown number of balls, and occasionally an eight-mark question combining several parts on one situation. For every part, write the sample space (or its size), the favourable outcomes (listed when small), and the fraction, then simplify. If a part asks 'is this experiment fair?' answer with the probabilities of each player winning and compare them; a game is fair when the players have equal probabilities of winning.

Quick checks. Do the probabilities of all elementary events add to 1? Does P(E) + P(not E) = 1? Is the answer between 0 and 1? Does the answer make sense (a very likely event should have a probability near 1)? Two minutes spent on these checks saves many marks.

Verifying by experiment. A good habit is to test a theoretical answer with a quick simulation: toss a coin twenty times, roll a die, draw cards. Seeing the experimental proportion hover near the calculated value builds confidence in the formula and in the relation between the two kinds of probability.