Class 9 Mathematics Chapter 7: The Mathematics of Maybe: Introduction to Probability

An event is random when all its possible outcomes are known but it cannot be predicted which of them will actually occur. Tossing a coin is an example: it will certainly land heads or tails, and nothing beyond that can be said about any one toss.

Every probability lies between 0 and 1 inclusive. A probability of 0 means the event is impossible and a probability of 1 means it is certain.

The experimental probability is $\dfrac{8}{50} = 0.16$, or 16 per cent. It does not claim that the die is loaded, and it does not predict how often a 4 will come up in any future set of rolls; it reports only what happened in those fifty trials.

The theoretical probability of an event is the number of favourable outcomes divided by the number of possible outcomes. The condition is that all those outcomes must be equally likely, as they are for a fair coin, an unbiased die or a well-shuffled pack.

There are 12 balls in all, and each is equally likely to be drawn. The probability of green is $\dfrac{5}{12}$ and the probability of red is $\dfrac{7}{12}$. As a check the two add to 1, as they must, since every draw gives one colour or the other.

Experimental probability is found from evidence. The experiment is carried out many times, or past records are examined, and the number of times the event occurred is divided by the total number of trials; the result is also called the relative frequency. Theoretical probability is found by reasoning instead: the number of favourable outcomes is divided by the number of possible outcomes, on the assumption that every outcome is equally likely. The experimental kind is used when no such assumption can be made or when only data is available; the theoretical kind is used when the set-up is known to be fair.

The two numbers are close but not equal, and there is no reason to expect them to be equal. Relative frequency wobbles about the theoretical value and settles down towards it only as the number of trials grows large; fifty rolls is a short run, and a result of eight fours rather than the eight or nine the theory suggests is entirely ordinary. Real evidence of bias would be a large gap that persisted over a great many trials, not a small one over fifty.

The relative frequency of a late bus is $\dfrac{35}{140} = \dfrac{1}{4}$, or 0.25, and that is the estimate. It should be treated with care, because an estimate is only as good as the evidence behind it. If those 140 buses were all recorded in one stretch of bad weather, or all at rush hour, the figure may say very little about the next bus on an ordinary day.

The theoretical formula divides favourable outcomes by possible outcomes, and it is valid only when all those outcomes are equally likely. Sectors of different sizes are not equally likely, since the pointer has more room to stop in a large one, so simply writing one over four would be wrong. What can be done instead is to spin the spinner a great many times and use the relative frequency of each sector as an experimental estimate. The alternative is to work from the angles of the sectors, taking each probability as its share of the full turn.

A probability is a count of favourable cases against a count of all the cases, or of occurrences against trials. Neither count can be negative, so the fraction cannot be negative either, and the smallest it can be is 0, when nothing is favourable. Nor can the favourable cases outnumber all the cases, so the fraction can never exceed 1, and it reaches 1 only when every case is favourable. Those two extremes are exactly what impossible and certain mean.

(i) Probability measures how likely an event is to happen. It is a measurement like length or area, but of likelihood rather than of a physical quantity.

(ii) A random event is one whose possible outcomes are all known but where it cannot be said in advance which of them will occur.

(iii) Every probability lies between 0 and 1 inclusive. A value of 0 means the event is impossible and a value of 1 means it is certain; 0.5 means an even chance.

(iv) Experimental probability, from evidence of what has happened, and theoretical probability, from reasoning about equally likely outcomes.

(v) A subjective probability is one person's judgement of likelihood from how they read the evidence. It is not used here because two people may read the same evidence differently and reach different numbers, whereas an objective estimate is one that anybody following the same method would reach.

(i) It is the number of times the event occurred divided by the total number of trials, and it is also called the relative frequency.

(ii) It is the number of favourable outcomes divided by the number of possible outcomes, on condition that all the outcomes are equally likely.

(iii) $\dfrac{108}{200} = 0.54$.

(iv) $\dfrac{1}{2}$, or 0.5.

(v) The two differ by only 0.04, which is what a run of 200 tosses would naturally produce. Relative frequency wobbles about the theoretical value and approaches it only as the number of trials grows large. The difference is therefore no evidence at all that the coin is unfair.

(i) $\dfrac{35}{140} = \dfrac{1}{4}$, or 0.25.

(ii) The other 105 deliveries were on time, so the estimate is $\dfrac{105}{140} = \dfrac{3}{4}$, or 0.75.

(iii) They must add to 1. Every delivery is either late or on time, so between them the two events cover every case and account for the whole of the record.

(iv) The record may not be representative. If those 140 deliveries all fell in one busy season, or all came from one supplier, the figure may say little about the next delivery under different conditions.

(v) By collecting more records, and by making sure they are spread across different times, seasons and suppliers, so that the relative frequency settles down and is not distorted by any one set of circumstances.

(i) $P(4) = \dfrac{1}{6}$, since one of six faces is wanted. $P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$, since 2, 4 and 6 qualify. $P(\text{above } 4) = \dfrac{2}{6} = \dfrac{1}{3}$, since only 5 and 6 qualify.

(ii) 0. No face of the die shows a 7, so the event has no favourable outcome at all.

(iii) 1. Every face shows a number below 7, so the event is certain.

(iv) They illustrate the two ends of the probability scale: 0 for something impossible and 1 for something certain. Every other event of this experiment falls strictly between them.

(v) A probability is a count of favourable cases over a count of all cases, and the favourable ones are a part of the whole, so they can never outnumber it. A value above 1 means either that the two counts were divided the wrong way round or that something was counted twice.

A sample space is the complete list of all possible outcomes of an experiment, written $S$. Every possible outcome must appear on it, and no outcome may appear more than once.

The sample space is $S = {HH, HT, TH, TT}$, and its size is $n(S) = 4$. The outcomes $HT$ and $TH$ are different, because the two coins can be told apart.

An event is a subset of the sample space, that is, any outcome or collection of outcomes drawn from the list. For one die with $S = {1, 2, 3, 4, 5, 6}$, the event "greater than 4" is $E = {5, 6}$.

The probability is $P(E) = \dfrac{n(E)}{n(S)}$, the number of elements in the event divided by the number in the sample space. It holds when all the outcomes in the sample space are equally likely.

The sample space is ${HH, HT, TH, TT}$, so $n(S) = 4$. The event is $E = {HH, HT, TH}$, so $n(E) = 3$, and the probability is $\dfrac{3}{4}$.

It is tempting to treat "one head and one tail" as a single outcome and write ${HH, HT, TT}$. But the two coins can be told apart, and getting a head on the first with a tail on the second is a different result from the other way about. Keeping them separate also matters for the arithmetic. If they were merged, the merged outcome would happen twice as often as $HH$, so the listed outcomes would no longer be equally likely and the formula $\dfrac{n(E)}{n(S)}$ would give wrong answers.

Each of the six faces can be paired with either result of the coin, so $n(S) = 6 \times 2 = 12$. The outcomes with an even number and heads are $2H$, $4H$ and $6H$, so $n(E) = 3$ and the probability is $\dfrac{3}{12} = \dfrac{1}{4}$.

"At least one head" means one or more, so it includes the outcome with two heads: $E = {HH, HT, TH}$ and the probability is $\dfrac{3}{4}$. "Exactly one head" means one and no more, so the two-head outcome is excluded: $E = {HT, TH}$ and the probability is $\dfrac{2}{4} = \dfrac{1}{2}$. The two phrases pick out different subsets of the same sample space.

The rule is $P(\text{not } E) = 1 - P(E)$. It holds because the outcomes in $E$ and those outside it together make up the whole sample space with nothing left over, so their probabilities must add to 1. For two coins, the probability of no head at all is $\dfrac{1}{4}$, since only $TT$ qualifies, so the probability of at least one head is $1 - \dfrac{1}{4} = \dfrac{3}{4}$ — reached by counting one outcome instead of three.

With that sample space there are two outcomes, and the formula $\dfrac{n(E)}{n(S)}$ would give $\dfrac{1}{2}$ for green. But the formula requires the outcomes to be equally likely, and green and red are not: there are more red balls, so red is the more likely of the two. Listing the twelve individual balls instead restores equal likelihood, since any one ball is as likely to be drawn as any other, and then five of the twelve are green, giving $\dfrac{5}{12}$.

(i) A sample space is the complete list of all possible outcomes of an experiment, written $S$. Every possible outcome must appear on it, and no outcome may appear more than once.

(ii) The sample size is the number of elements in the sample space, written $n(S)$.

(iii) For one die, $S = {1, 2, 3, 4, 5, 6}$ with $n(S) = 6$. For two coins, $S = {HH, HT, TH, TT}$ with $n(S) = 4$.

(iv) An event is a subset of the sample space: any outcome, or collection of outcomes, drawn from that list.

(v) $P(E) = \dfrac{n(E)}{n(S)}$, valid when all the outcomes in the sample space are equally likely.

(i) Four: ${HH, HT, TH, TT}$.

(ii) Eight, since each coin doubles the count: $2 \times 2 \times 2 = 8$.

(iii) Twelve, since each of six faces goes with either of two coin results: $6 \times 2 = 12$.

(iv) Thirty-six, since $6 \times 6 = 36$.

(v) When two or more things happen together, every outcome of the first can be paired with every outcome of the second, so the counts multiply. This is why adding a further coin doubles the total and why two dice give the square of six rather than twice six.

(i) $n(S) = 6 \times 6 = 36$.

(ii) The pairs totalling 7 are $(1,6)$, $(2,5)$, $(3,4)$, $(4,3)$, $(5,2)$ and $(6,1)$, six in all, so the probability is $\dfrac{6}{36} = \dfrac{1}{6}$.

(iii) Only $(1,1)$ gives a total of 2, so the probability is $\dfrac{1}{36}$.

(iv) The doubles are $(1,1)$ up to $(6,6)$, six in all, so the probability is $\dfrac{6}{36} = \dfrac{1}{6}$.

(v) Because there are six different ways of making 7 and only one way of making 2. Each individual pair is equally likely, so the total that can be reached by more pairs is the more likely one.

(i) $P(\text{not } E) = 1 - P(E)$.

(ii) The outcomes in $E$ and those not in $E$ between them make up the whole sample space, with nothing left over and nothing counted twice. Since the probabilities of all the outcomes add to 1, the two events' probabilities must add to 1 as well.

(iii) The only outcome with no head is $TT$, with probability $\dfrac{1}{4}$, so at least one head has probability $1 - \dfrac{1}{4} = \dfrac{3}{4}$.

(iv) $\dfrac{5}{12}$, since five of the twelve equally likely balls are green.

(v) That sample space has two outcomes, so the formula would give $\dfrac{1}{2}$. But the formula needs the outcomes to be equally likely, and green and red are not: there are more red balls than green ones, so red is the likelier of the two.

A multi-step experiment is one carried out in stages, each stage being a trial of its own, so that the outcome of the whole thing is the combination of what happened at every stage. Tossing a coin twice is an example: the first toss is one stage and the second another, and the result is a pair such as $HT$.

Begin at a single point and draw one branch for each possible outcome of the first stage. From the end of every one of those branches draw a fresh branch for each outcome of the second stage, and carry on with one fan of branches per stage. A complete route from the starting point to a tip represents one outcome of the whole experiment, and reading the labels along it in order gives that outcome.

The number of paths is the product of the numbers of outcomes at each stage. For three coin tosses each stage has two outcomes, so there are $2 \times 2 \times 2 = 8$ paths.

Each branch of a fair coin carries a probability of $\dfrac{1}{2}$, and the probability of a complete path is the product of its branches. So $P(HHH) = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$.

They must add up to exactly 1. If they do not, then either a path has been left off the tree or one of the branch probabilities has been written down wrongly.

The eight paths through the tree are $HHH$, $HHT$, $HTH$, $HTT$, $THH$, $THT$, $TTH$ and $TTT$, and each has probability $\dfrac{1}{8}$. Exactly two heads means two and no more, so the qualifying paths are $HHT$, $HTH$ and $THH$, three of them. The probability is therefore $\dfrac{3}{8}$.

Every outcome of the first stage can be followed by every outcome of the second, so each of the first stage's branches splits into a full set of the second stage's branches; that is a multiplication rather than an addition. With two coins the difference is hidden, because $2 \times 2$ and $2 + 2$ both come to 4. With three coins it shows plainly: the true count is $2 \times 2 \times 2 = 8$, while adding would give only 6.

The coin gives two branches and each of them fans out into the six faces of the die, so there are $2 \times 6 = 12$ paths in all. They run $H1$ to $H6$ and $T1$ to $T6$. Heads with an even number is satisfied by $H2$, $H4$ and $H6$, three paths out of twelve, so the probability is $\dfrac{3}{12} = \dfrac{1}{4}$.

If the first stage were drawn as just apple and orange there would be two branches, but they would not be equally likely, since two of the three fruits in basket A are oranges. Drawing all three fruits separately keeps every path as likely as every other. That gives $3 \times 2 = 6$ paths. An orange with a mango happens along two of them, one for each orange, so the probability is $\dfrac{2}{6} = \dfrac{1}{3}$.

Multiplying along a path is valid when the stages are independent, that is, when what happens at one stage has no effect on what happens at the next; separate tosses of a coin are independent in this way. A tree remains worth drawing because multiplying tells you only how many paths there are, not which ones they are. Any question about a named event, such as exactly two heads, needs the paths themselves so that the qualifying ones can be picked out and counted.

(i) A multi-step experiment is carried out in stages, each stage a trial of its own, with the outcome of the whole being the combination of what happened at every stage.

(ii) Start from one point and draw a branch for each outcome of the first stage. From the end of every branch, fan out a branch for each outcome of the next stage, and continue with one fan per stage.

(iii) One complete route from the starting point to a tip is one outcome of the whole experiment, named by reading its labels in order.

(iv) By multiplying the numbers of outcomes at the separate stages.

(v) Four tosses of a coin give $2 \times 2 \times 2 \times 2 = 16$ paths.

(i) $HHH$, $HHT$, $HTH$, $HTT$, $THH$, $THT$, $TTH$, $TTT$.

(ii) Each has probability $\dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{8}$, and the eight together add to 1.

(iii) The paths with exactly two heads are $HHT$, $HTH$ and $THH$, so the probability is $\dfrac{3}{8}$.

(iv) Seven of the eight paths contain a head, since only $TTT$ does not, so the probability is $\dfrac{7}{8}$.

(v) Take the complement. The probability of no head at all is $\dfrac{1}{8}$, so the probability of at least one head is $1 - \dfrac{1}{8} = \dfrac{7}{8}$, reached by counting one path instead of seven.

(i) Three fruits in basket A and two in basket B give $3 \times 2 = 6$ paths.

(ii) If the first stage were drawn as apple and orange, there would be two branches but they would not be equally likely, since two of the three fruits are oranges. The counting formula needs every path to be as likely as every other, so all three fruits are listed.

(iii) Four of the six paths begin with an orange, so the probability is $\dfrac{4}{6} = \dfrac{2}{3}$.

(iv) Three of the six paths end with the mango, so the probability is $\dfrac{3}{6} = \dfrac{1}{2}$.

(v) Just one path has the apple with the mango, so the probability is $\dfrac{1}{6}$.

(i) By multiplying together the probabilities written along its branches.

(ii) The stages must be independent, that is, what happens at one stage must not affect what happens at the next. Separate tosses of a coin satisfy this.

(iii) The probabilities of all the complete paths must add up to exactly 1.

(iv) That either a path has been left off the tree, so that not every outcome is represented, or one of the branch probabilities has been written down wrongly. It is a cheap check and it catches an incomplete fan of branches at once.

(v) Six tosses give $2^6 = 64$ paths, and the drawing grows faster than the page. The sensible course is to trust the multiplication for the count, and to draw a tree only when the list of outcomes itself is needed and small enough to write.

An event is random when all its possible outcomes are known but it cannot be predicted which of them will actually occur. Tossing a coin is an example: it will certainly land heads or tails, and nothing beyond that can be said about any one toss.

Every probability lies between 0 and 1 inclusive. A probability of 0 means the event is impossible and a probability of 1 means it is certain.

The experimental probability is $\dfrac{8}{50} = 0.16$, or 16 per cent. It does not claim that the die is loaded, and it does not predict how often a 4 will come up in any future set of rolls; it reports only what happened in those fifty trials.

A sample space is the complete list of all possible outcomes of an experiment, written $S$. Every possible outcome must appear on it, and no outcome may appear more than once.

The sample space is $S = {HH, HT, TH, TT}$, and its size is $n(S) = 4$. The outcomes $HT$ and $TH$ are different, because the two coins can be told apart.

An event is a subset of the sample space, that is, any outcome or collection of outcomes drawn from the list. For one die with $S = {1, 2, 3, 4, 5, 6}$, the event "greater than 4" is $E = {5, 6}$.

A multi-step experiment is one carried out in stages, each stage being a trial of its own, so that the outcome of the whole thing is the combination of what happened at every stage. Tossing a coin twice is an example: the first toss is one stage and the second another, and the result is a pair such as $HT$.

Begin at a single point and draw one branch for each possible outcome of the first stage. From the end of every one of those branches draw a fresh branch for each outcome of the second stage, and carry on with one fan of branches per stage. A complete route from the starting point to a tip represents one outcome of the whole experiment, and reading the labels along it in order gives that outcome.

The number of paths is the product of the numbers of outcomes at each stage. For three coin tosses each stage has two outcomes, so there are $2 \times 2 \times 2 = 8$ paths.