NCERT Solutions · Class 9 Mathematics The Mathematics of Maybe: Introduction to Probability

Relative frequency from a sample. Nobody can count a whole bag of sweets one by one, so instead we count a sample and read off the fraction of each colour in it. That fraction - the relative frequency - is our estimate of the probability for the whole bag.First check that the sample really is $\displaystyle 30$ sweets: \[10+8+7+5=30 \quad\checkmark \](i) Probability of green. In the sample, $\displaystyle 8$ of the $\displaystyle 30$ sweets are green, and each of the $\displaystyle 30$ is equally likely to be the one picked: \[P(\text{green})=\frac{\text{number of green sweets}}{\text{total number of sweets}}=\frac{8}{30}=\frac{4}{15}\approx 0.27 \](ii) Estimating the yellow sweets in the whole bag. The yellow relative frequency in the sample is \[\frac{7}{30} \] If the bag as a whole has roughly the same mix as the sample, then about \(\displaystyle \tfrac{7}{30}\) of the $\displaystyle 600$ sweets are yellow: \[\frac{7}{30}\times 600 = 7\times 20 = 140 \]A second way to see the same thing: \(\displaystyle 600\div 30=20\), so the bag is $\displaystyle 20$ "sample-loads" big, and each yellow sweet counted in the sample stands for $\displaystyle 20$ yellow sweets in the bag - \(\displaystyle 7\times 20=140\). Both routes agree.This is an estimate, not an exact count. If the teacher scooped out a different $\displaystyle 30$ sweets she would probably find $\displaystyle 6$ or $\displaystyle 8$ yellow ones instead of $\displaystyle 7$, and the estimate would change. A larger sample would make it more trustworthy.Answers: (i) \(\displaystyle P(\text{green})=\dfrac{4}{15}\approx 0.27\); (ii) about \(\displaystyle 140\) yellow sweets.

Scaling a sample up to the whole school. The $\displaystyle 40$ students surveyed stand in for all 800. Whatever fraction of the sample chose a club, we take that as our estimate of the fraction of the whole school that would choose it.Check the sample total first: \[14+11+9+6=40 \quad\checkmark \](i) Probability of the Arts Club. Of the $\displaystyle 40$ students in the sample, $\displaystyle 11$ chose Arts, and any one of the $\displaystyle 40$ is equally likely to be picked: \[P(\text{Arts})=\frac{11}{40}=0.275 \](ii) Estimating the Sports Club numbers in the school. In the sample, \(\displaystyle \dfrac{9}{40}\) of the students prefer Sports. Applying the same fraction to all $\displaystyle 800$ students: \[\frac{9}{40}\times 800 = 9\times 20 = 180 \]Checking it another way: \(\displaystyle 800\div 40=20\), so each student surveyed represents $\displaystyle 20$ students in the school, and \(\displaystyle 9\times 20=180\).A useful check on the whole answer: scaling all four clubs up by the factor $\displaystyle 20$ gives \(\displaystyle 280+220+180+120=800\), which is exactly the school roll - so the four estimates are consistent with one another.Remember this is only an estimate. It is reliable only if the $\displaystyle 40$ students really were chosen at random; if the survey had been taken outside the sports field, the Sports Club figure would come out far too high.Answers: (i) \(\displaystyle P(\text{Arts})=\dfrac{11}{40}=0.275\); (ii) about \(\displaystyle 180\) students.

An experiment you perform - your numbers will be your own. This question asks you to actually toss a coin $\displaystyle 20$ times, so there is no single correct set of answers for (i), (ii) and (iii): every honest record is correct. What is fixed is the method, and the answer to (iv). Below is one real record of $\displaystyle 20$ tosses, worked through in full, so you can see what to do with yours.A specimen record. Tossing a coin $\displaystyle 20$ times gave, in order: \[\mathrm{T\ T\ H\ T\ H\ H\ H\ H\ T\ T\ H\ T\ H\ H\ H\ H\ T\ T\ H\ T} \]

Outcome	Tally count	Relative frequency
Heads	$\displaystyle 11$	\(\displaystyle \frac{11}{20}=0.55\)
Tails	$\displaystyle 9$	\(\displaystyle \frac{9}{20}=0.45\)
Total	$\displaystyle 20$	\(\displaystyle 1\)

(i) Heads came up \(\displaystyle 11\) times. (Count your own H's; whatever you get is your answer.)(ii) Tails came up \(\displaystyle 9\) times. Your two counts must add to $\displaystyle 20$ - that is the check that you have not miscounted: \[11+9=20 \quad\checkmark \](iii) Experimental probability of heads. \[P(\text{heads})=\frac{\text{number of heads}}{\text{number of tosses}}=\frac{11}{20}=0.55 \] For your own record, divide your number of heads by $\displaystyle 20$ in the same way. Notice that \(\displaystyle 0.55\) is near \(\displaystyle 0.5\) but not equal to it - with only $\displaystyle 20$ tosses that is exactly what you should expect.(iv) The probability of tails on one more toss. For a fair coin this is \[P(\text{tails})=\frac{1}{2} \] and this is the same whatever your first $\displaystyle 20$ tosses happened to be. A coin has no memory: it does not know that $\displaystyle 11$ heads have already come up, and it cannot "make up" for them. The 21st toss is a fresh experiment with two equally likely outcomes.(If instead you are asked to estimate the chance of tails from your data alone, you would quote your experimental value \(\displaystyle \tfrac{9}{20}=0.45\) - close to \(\displaystyle 0.5\), and it would creep closer still if you tossed $\displaystyle 200$ times.)Answers: (i)-(iii) depend on your own tosses - in the specimen record above, $\displaystyle 11$ heads, $\displaystyle 9$ tails, and \(\displaystyle P(\text{heads})=\tfrac{11}{20}=0.55\). (iv) \(\displaystyle P(\text{tails})=\tfrac{1}{2}\) for a fair coin, no matter what the earlier tosses were.

When there is no theory to fall back on - measure. With a coin or a die you can reason out the probabilities in advance, because the faces are symmetric and so equally likely. A paper cup is not symmetric: it is light, open at one end and has a wide curved side. There is no way to argue from the shape that the three resting positions - on its bottom, upside down on its top, or on its side - are equally likely, and in fact they are not. So the only way to assign probabilities here is to toss it many times and use the relative frequency of each outcome: \[P(\text{outcome})=\frac{\text{number of times that outcome occurred}}{\text{total number of tosses}} \]Your results will be your own. The cup you use, and how you toss it, will change the numbers, so there is no single correct answer. Here is one complete set of results from $\displaystyle 100$ tosses of a paper cup, to show the working.

Resting position	Frequency (out of $\displaystyle 100$)	Experimental probability
On its side	$\displaystyle 62$	\(\displaystyle \frac{62}{100}=0.62\)
On its bottom	$\displaystyle 27$	\(\displaystyle \frac{27}{100}=0.27\)
Upside down on its top	$\displaystyle 11$	\(\displaystyle \frac{11}{100}=0.11\)
Total	$\displaystyle 100$	\(\displaystyle 1.00\)

Two checks worth doing every time. Reading the result. In this run the cup landed on its side more than half the time - the side is the largest, most stable surface - and it very rarely balanced upside down on its narrow top. Your figures should show the same general pattern even though the exact numbers differ. Repeat the $\displaystyle 100$ tosses and you will not get $\displaystyle 62$ again; but with $\displaystyle 200$ or $\displaystyle 500$ tosses the fractions settle down, and that is what makes them worth calling probabilities.Answer: assign each outcome the fraction \(\displaystyle \dfrac{\text{its frequency}}{100}\). In the specimen run above: \(\displaystyle P(\text{side})=0.62\), \(\displaystyle P(\text{bottom})=0.27\), \(\displaystyle P(\text{upside down on top})=0.11\), which sum to \(\displaystyle 1\). Your own three numbers will differ, and any set that comes from an honestly recorded $\displaystyle 100$ tosses and sums to \(\displaystyle 1\) is correct.

Counting favourable outcomes out of equally likely ones. A die is called fair precisely because its six faces are equally likely, and that is what lets us compute the probability by counting instead of by experimenting: \[P(E)=\frac{\text{number of outcomes in }E}{\text{total number of equally likely outcomes}} \]The sample space - every outcome the roll can produce - is \[S=\{1,\,2,\,3,\,4,\,5,\,6\},\qquad n(S)=6 \]The event "an even number" collects the outcomes divisible by $\displaystyle 2$: \[E=\{2,\,4,\,6\},\qquad n(E)=3 \]Therefore \[P(\text{even})=\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2} \]A check. The odd numbers are \(\displaystyle \{1,3,5\}\), also three of them, so \(\displaystyle P(\text{odd})=\tfrac{3}{6}=\tfrac{1}{2}\) as well. Even and odd between them cover every face, and indeed \(\displaystyle \tfrac12+\tfrac12=1\) - as the probabilities of an event and its opposite must.Answer: \(\displaystyle P(\text{even number})=\dfrac{3}{6}=\dfrac{1}{2}\).

Two different kinds of probability, and what joins them. Experimental probability is measured - you count what actually happened. Theoretical probability is reasoned out from the symmetry of the die before you roll it at all. They are answers to slightly different questions, which is why they need not match.(i) Experimental probability of a '$\displaystyle 3$'. The '$\displaystyle 3$' appeared $\displaystyle 3$ times in $\displaystyle 12$ rolls: \[P_{\text{exp}}(3)=\frac{\text{number of 3s obtained}}{\text{number of rolls}}=\frac{3}{12}=\frac{1}{4}=0.25 \](ii) Theoretical probability of a '$\displaystyle 3$'. A fair die has six equally likely faces, \(\displaystyle S=\{1,2,3,4,5,6\}\), and exactly one of them is a $\displaystyle 3$: \[P_{\text{theo}}(3)=\frac{1}{6}\approx 0.167 \](iii) Why they differ, and what more rolls would do. The gap is not a mistake in either calculation - it is the ordinary randomness of a short experiment. With only $\displaystyle 12$ rolls, the "expected" number of 3s is \(\displaystyle \tfrac{1}{6}\times 12 = 2\). Getting $\displaystyle 3$ instead of $\displaystyle 2$ is a difference of just one extra roll landing on $\displaystyle 3$, which is entirely unremarkable. But that one extra roll shifts the fraction a long way, from \(\displaystyle \tfrac{2}{12}=0.167\) to \(\displaystyle \tfrac{3}{12}=0.25\), because the denominator is so small.As the number of rolls grows, a stray roll or two stops mattering, and the experimental fraction settles closer and closer to \(\displaystyle \tfrac{1}{6}\). This is the key idea of the chapter: experimental probability approaches theoretical probability as the number of trials increases.

Number of rolls	Expected number of 3s	What a typical run might give
$\displaystyle 12$	$\displaystyle 2$	anywhere from $\displaystyle 0$ to $\displaystyle 5$ - so \(\displaystyle P_{\text{exp}}\) from \(\displaystyle 0\) to about \(\displaystyle 0.42\)
$\displaystyle 60$	$\displaystyle 10$	roughly $\displaystyle 6$ to $\displaystyle 14$, so \(\displaystyle P_{\text{exp}}\) near \(\displaystyle 0.10\) to \(\displaystyle 0.23\)
$\displaystyle 600$	$\displaystyle 100$	roughly $\displaystyle 85$ to $\displaystyle 115$, so \(\displaystyle P_{\text{exp}}\) near \(\displaystyle 0.14\) to \(\displaystyle 0.19\)
$\displaystyle 6000$	$\displaystyle 1000$	roughly $\displaystyle 950$ to $\displaystyle 1050$, so \(\displaystyle P_{\text{exp}}\) near \(\displaystyle 0.158\) to \(\displaystyle 0.175\)

(The expected numbers come from \(\displaystyle \tfrac{1}{6}\times 12=2\), \(\displaystyle \tfrac{1}{6}\times 60=10\), \(\displaystyle \tfrac{1}{6}\times 600=100\), \(\displaystyle \tfrac{1}{6}\times 6000=1000\).)Notice what is not being claimed. Rolling $\displaystyle 6000$ times will not give exactly $\displaystyle 1000$ threes; in fact you will typically be off by a few dozen. The count drifts further from its expected value as you roll more - it is the fraction that closes in on \(\displaystyle \tfrac16\), because the same-sized error is being divided by a much bigger number.Answers: (i) \(\displaystyle P_{\text{exp}}(3)=\dfrac{3}{12}=\dfrac{1}{4}\); (ii) \(\displaystyle P_{\text{theo}}(3)=\dfrac{1}{6}\); (iii) they differ only because $\displaystyle 12$ rolls is far too short a run for chance variation to average out - with $\displaystyle 60$, $\displaystyle 600$ and $\displaystyle 6000$ rolls you would expect about $\displaystyle 10$, $\displaystyle 100$ and $\displaystyle 1000$ threes, and the experimental probability would come steadily closer to \(\displaystyle \dfrac{1}{6}\).

The sample space is the list of everything that can happen. For one roll of a $\displaystyle 6$-sided die, the possible results are the numbers printed on the six faces, and no others - the die must come to rest showing exactly one face. \[S=\{1,\,2,\,3,\,4,\,5,\,6\} \] Counting the members of that list: \[n(S)=6 \]Note that the question asks how many outcomes there are, not how likely each is. The count would still be $\displaystyle 6$ even for a loaded die whose faces were not equally likely - the sample space records what is possible, and the probabilities are a separate layer of information laid on top of it.Answer: \(\displaystyle 6\) possible outcomes, namely \(\displaystyle S=\{1,2,3,4,5,6\}\).

Listing a sample space. The sample space \(\displaystyle S\) of an experiment is the set of all the outcomes it can produce, with nothing left out and nothing counted twice. When an experiment has two parts done together, each outcome has to record both parts.(i) Rolling a die and tossing a coin together. One outcome must say what the die showed and what the coin showed, so write it as an ordered pair (die, coin). The die contributes $\displaystyle 6$ possibilities and the coin $\displaystyle 2$, and every die result can pair with either coin result: \[S=\{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\} \] \[n(S)=6\times 2=12 \] The multiplication \(\displaystyle 6\times 2\) is just a shortcut for the counting you can see in the list: $\displaystyle 6$ outcomes with \(\displaystyle H\), then the same $\displaystyle 6$ again with \(\displaystyle T\).(ii) Choosing a random integer between \(\displaystyle -5\) and \(\displaystyle +5\). Read strictly, "between \(\displaystyle -5\) and \(\displaystyle +5\)" means the two end numbers are not themselves included, so the integers available are \[S=\{-4,-3,-2,-1,0,1,2,3,4\},\qquad n(S)=9 \] Do not forget \(\displaystyle 0\) - it is an integer, and it lies between \(\displaystyle -5\) and \(\displaystyle 5\). (A word of caution: the word "between" is used loosely in everyday speech. If your teacher intends the endpoints to be included, the sample space is \(\displaystyle S=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}\) with \(\displaystyle n(S)=11\). Say which reading you are using, and either is defensible; the strict reading, giving $\displaystyle 9$, is the one taken here.)(iii) Drawing one ball from a box of $\displaystyle 5$ green and $\displaystyle 7$ red balls. This depends on what you decide to record. If you only record the colour*, then \(\displaystyle S=\{\text{green},\ \text{red}\}\) and \(\displaystyle n(S)=2\). But be careful - these two outcomes are not equally likely, so you must not compute probabilities by dividing by 2. To get a sample space of equally likely* outcomes, treat the balls as distinguishable - imagine numbering them: \[S=\{G_1,G_2,G_3,G_4,G_5,\ R_1,R_2,R_3,R_4,R_5,R_6,R_7\},\qquad n(S)=5+7=12 \] Each of these $\displaystyle 12$ balls is equally likely to be the one drawn, and now counting works: \[P(\text{green})=\frac{5}{12},\qquad P(\text{red})=\frac{7}{12},\qquad \frac{5}{12}+\frac{7}{12}=1\ \checkmark \]Answers: (i) \(\displaystyle S=\{(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)\}\), \(\displaystyle n(S)=12\); (ii) \(\displaystyle S=\{-4,-3,-2,-1,0,1,2,3,4\}\), \(\displaystyle n(S)=9\) (or \(\displaystyle n(S)=11\) if \(\displaystyle -5\) and \(\displaystyle +5\) are meant to be included); (iii) by colour \(\displaystyle S=\{\text{green},\text{red}\}\) with $\displaystyle 2$ unequally likely outcomes, or, taking the balls as distinct, \(\displaystyle S=\{G_1,\dots,G_5,R_1,\dots,R_7\}\) with \(\displaystyle n(S)=12\) equally likely outcomes.

Pairing every choice with every other choice. A villager makes two independent choices - one snack out of $\displaystyle 3$, one drink out of $\displaystyle 2$ - and an outcome must record both. So each outcome is a pair (snack, drink), and every one of the $\displaystyle 3$ snacks can go with either of the $\displaystyle 2$ drinks.(i) The sample space. Take the snacks one at a time and give each its two drinks; that way nothing is missed and nothing is repeated:

Snack	With Chai	With Lassi
Samosa	(Samosa, Chai)	(Samosa, Lassi)
Pakora	(Pakora, Chai)	(Pakora, Lassi)
Bhaji	(Bhaji, Chai)	(Bhaji, Lassi)

Written as a set, \[S=\{(\text{Samosa},\text{Chai}),\ (\text{Samosa},\text{Lassi}),\ (\text{Pakora},\text{Chai}),\ (\text{Pakora},\text{Lassi}),\ (\text{Bhaji},\text{Chai}),\ (\text{Bhaji},\text{Lassi})\} \] \[n(S)=3\times 2=6 \] The table makes the multiplication visible: $\displaystyle 3$ rows, $\displaystyle 2$ columns, \(\displaystyle 3\times 2=6\) cells, and each cell is one outcome.(ii) The event "Selecting Samosa as a snack". An event is a part of the sample space - the collection of those outcomes in which the described thing happens. A person who chooses a samosa still has to drink something, so this event contains two outcomes, not one: \[E=\{(\text{Samosa},\text{Chai}),\ (\text{Samosa},\text{Lassi})\},\qquad n(E)=2 \] That is simply the first row of the table. If every combination were equally likely, this event would have probability \(\displaystyle \tfrac{2}{6}=\tfrac{1}{3}\) - which makes sense, since the samosa is one snack out of three.Answers: (i) \(\displaystyle S=\{(\text{Samosa},\text{Chai}),(\text{Samosa},\text{Lassi}),(\text{Pakora},\text{Chai}),(\text{Pakora},\text{Lassi}),(\text{Bhaji},\text{Chai}),(\text{Bhaji},\text{Lassi})\}\), with \(\displaystyle n(S)=6\); (ii) \(\displaystyle E=\{(\text{Samosa},\text{Chai}),(\text{Samosa},\text{Lassi})\}\), with \(\displaystyle n(E)=2\).

A tree diagram keeps the two picks separate. Basket A holds $\displaystyle 3$ fruits (one apple, two oranges) and Basket B holds $\displaystyle 2$ fruits (one banana, one mango). The crucial point is that the two oranges are two different fruits: your hand is equally likely to close on either one. So the tree must give them a branch each - otherwise the apple and "an orange" would look equally likely, and they are not.(i) The tree. Draw the first stage as $\displaystyle 3$ branches for Basket A. From the tip of each of those, draw $\displaystyle 2$ branches for Basket B. That gives \(\displaystyle 3\times 2=6\) endpoints, one for each possible pair. Written out branch by branch:

Pick from Basket A	Pick from Basket B	Pair at the end of the branch
Apple	Banana	(Apple, Banana)
Apple	Mango	(Apple, Mango)
Orange $\displaystyle 1$	Banana	(Orange $\displaystyle 1$, Banana)
Orange $\displaystyle 1$	Mango	(Orange $\displaystyle 1$, Mango)
Orange $\displaystyle 2$	Banana	(Orange $\displaystyle 2$, Banana)
Orange $\displaystyle 2$	Mango	(Orange $\displaystyle 2$, Mango)

Each of the $\displaystyle 3$ first-stage branches carries probability \(\displaystyle \tfrac{1}{3}\), and each of the $\displaystyle 2$ second-stage branches carries probability \(\displaystyle \tfrac{1}{2}\).(ii) The sample space. Reading the $\displaystyle 6$ endpoints off the tree: \[S=\{(A,B),(A,M),(O_1,B),(O_1,M),(O_2,B),(O_2,M)\},\qquad n(S)=6 \] where \(\displaystyle A\) is the apple, \(\displaystyle O_1,O_2\) the two oranges, \(\displaystyle B\) the banana and \(\displaystyle M\) the mango. These $\displaystyle 6$ outcomes are equally likely, which is what makes them useful.(If you record only the kinds of fruit, the list shrinks to four: \(\displaystyle \{(\text{apple},\text{banana}),(\text{apple},\text{mango}),(\text{orange},\text{banana}),(\text{orange},\text{mango})\}\). But those four are not equally likely - each orange pair covers two branches of the tree and so is twice as likely as the matching apple pair. Never divide by $\displaystyle 4$ here.)(iii) Probability of one apple and one banana. Exactly one of the $\displaystyle 6$ equally likely outcomes is \(\displaystyle (A,B)\): \[P(\text{apple and banana})=\frac{1}{6} \]Check by multiplying along the branch. The chance of the apple from Basket A is \(\displaystyle \tfrac{1}{3}\) ($\displaystyle 1$ apple out of $\displaystyle 3$ fruits), and the chance of the banana from Basket B is \(\displaystyle \tfrac{1}{2}\) ($\displaystyle 1$ banana out of $\displaystyle 2$). The two picks are from different baskets, so neither affects the other, and the branch probability is their product: \[\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}\ \checkmark \] The two methods agree. As a further check, all $\displaystyle 6$ endpoints carry \(\displaystyle \tfrac16\) and \(\displaystyle 6\times\tfrac16=1\).Answers: (i) a tree with $\displaystyle 3$ branches (Apple, Orange $\displaystyle 1$, Orange $\displaystyle 2$) each splitting into $\displaystyle 2$ (Banana, Mango), giving $\displaystyle 6$ endpoints; (ii) \(\displaystyle S=\{(A,B),(A,M),(O_1,B),(O_1,M),(O_2,B),(O_2,M)\}\), \(\displaystyle n(S)=6\); (iii) \(\displaystyle P(\text{apple and banana})=\dfrac{1}{6}\).

Two picks, and the box is the same both times. The box holds \[3+4+2=9 \text{ pens} \] Because you put your pen back before your friend picks, your friend faces exactly the same box of $\displaystyle 9$ pens that you did. The two picks are therefore independent - what you drew tells us nothing about what your friend will draw - and both picks have the same colour probabilities: \[P(\text{red})=\frac{3}{9}=\frac13,\qquad P(\text{black})=\frac{4}{9},\qquad P(\text{green})=\frac{2}{9},\qquad \frac39+\frac49+\frac29=1\ \checkmark \](i) The possible outcomes, and the tree. An outcome records both colours in order (yours first, your friend's second). Your pick has $\displaystyle 3$ possible colours, and from each of those your friend's pick again has the same $\displaystyle 3$, so the tree has \(\displaystyle 3\times 3 = 9\) endpoints:

Your pen	Friend's pen	Probability of the branch
Red \(\displaystyle \left(\frac{3}{9}\right)\)	Red \(\displaystyle \left(\frac{3}{9}\right)\)	\(\displaystyle \frac{3}{9}\times\frac{3}{9}=\frac{9}{81}\)
Red \(\displaystyle \left(\frac{3}{9}\right)\)	Black \(\displaystyle \left(\frac{4}{9}\right)\)	\(\displaystyle \frac{12}{81}\)
Red \(\displaystyle \left(\frac{3}{9}\right)\)	Green \(\displaystyle \left(\frac{2}{9}\right)\)	\(\displaystyle \frac{6}{81}\)
Black \(\displaystyle \left(\frac{4}{9}\right)\)	Red \(\displaystyle \left(\frac{3}{9}\right)\)	\(\displaystyle \frac{12}{81}\)
Black \(\displaystyle \left(\frac{4}{9}\right)\)	Black \(\displaystyle \left(\frac{4}{9}\right)\)	\(\displaystyle \frac{16}{81}\)
Black \(\displaystyle \left(\frac{4}{9}\right)\)	Green \(\displaystyle \left(\frac{2}{9}\right)\)	\(\displaystyle \frac{8}{81}\)
Green \(\displaystyle \left(\frac{2}{9}\right)\)	Red \(\displaystyle \left(\frac{3}{9}\right)\)	\(\displaystyle \frac{6}{81}\)
Green \(\displaystyle \left(\frac{2}{9}\right)\)	Black \(\displaystyle \left(\frac{4}{9}\right)\)	\(\displaystyle \frac{8}{81}\)
Green \(\displaystyle \left(\frac{2}{9}\right)\)	Green \(\displaystyle \left(\frac{2}{9}\right)\)	\(\displaystyle \frac{4}{81}\)

So the outcomes are \(\displaystyle RR,\ RB,\ RG,\ BR,\ BB,\ BG,\ GR,\ GB,\ GG\) - nine of them. Note that \(\displaystyle RB\) and \(\displaystyle BR\) are different outcomes here, because the tree records who picked what.The nine branch probabilities add to \(\displaystyle \frac{9+12+6+12+16+8+6+8+4}{81}=\frac{81}{81}=1\), which confirms no branch has been missed. Notice also that these nine outcomes are not equally likely - black is the commonest colour in the box, so \(\displaystyle BB\) is the fattest branch.(ii) Probability that both pens are the same colour. "Same colour" means the branch ends in \(\displaystyle RR\), \(\displaystyle BB\) or \(\displaystyle GG\). These are three separate branches, so add their probabilities: \[P(\text{same colour})=\frac{9}{81}+\frac{16}{81}+\frac{4}{81}=\frac{29}{81}\approx 0.36 \]Check by counting pens instead of colours. Imagine the $\displaystyle 9$ pens all labelled. Your pick can be any of $\displaystyle 9$ pens and your friend's can be any of $\displaystyle 9$, giving \(\displaystyle 9\times 9=81\) equally likely pen-pairs. The pairs that match in colour number \[3\times 3+4\times 4+2\times 2=9+16+4=29 \] so \(\displaystyle P=\tfrac{29}{81}\) - the same answer by a completely different route.It follows that \(\displaystyle P(\text{different colours})=1-\tfrac{29}{81}=\tfrac{52}{81}\approx 0.64\): the pens are more likely to differ than to match.Answers: (i) $\displaystyle 9$ outcomes - \(\displaystyle RR, RB, RG, BR, BB, BG, GR, GB, GG\) - from a tree with $\displaystyle 3$ first-stage branches each splitting into the same $\displaystyle 3$; (ii) \(\displaystyle P(\text{same colour})=\dfrac{9}{81}+\dfrac{16}{81}+\dfrac{4}{81}=\dfrac{29}{81}\approx 0.36\).

The four facts that fix the probability scale.(i) An impossible event contains no outcomes at all - there is nothing in the sample space that makes it happen - so the count on top of the fraction is \(\displaystyle 0\): \[P(\text{impossible event})=\frac{0}{n(S)}=\mathbf{0} \](ii) The set of all possible outcomes of a random experiment is called the sample space, usually written \(\displaystyle S\). Every event is a subset of it.(iii) An event certain to happen contains every outcome in the sample space, so the fraction is the whole of itself: \[P(\text{certain event})=\frac{n(S)}{n(S)}=\mathbf{1} \](iv) A fair coin has \(\displaystyle S=\{H,T\}\) with the two outcomes equally likely, and heads is one of them: \[P(\text{heads})=\frac{1}{2}=\mathbf{0.5} \]Together (i) and (iii) are the two ends of the scale, which is why every probability satisfies \(\displaystyle 0\le P(E)\le 1\).Answers: (i) \(\displaystyle 0\); (ii) sample space; (iii) \(\displaystyle 1\); (iv) \(\displaystyle \dfrac{1}{2}\) (i.e. \(\displaystyle 0.5\)).

Frequency counts; relative frequency compares. The frequency of an outcome is the plain number of times it occurred. The relative frequency is that number divided by the total, so it tells you what share of the whole group it is - and it is the relative frequency, not the frequency, that estimates a probability.Here the frequency is given: $\displaystyle 15$ students like football. What the sentence is asking you to supply is the relative frequency: \[\text{relative frequency}=\frac{\text{number who like football}}{\text{total number surveyed}}=\frac{15}{50}=\frac{3}{10}=0.3 \](The fraction \(\displaystyle \tfrac{15}{50}\) reduces by dividing top and bottom by 5.)You can also read this as a percentage: \(\displaystyle 0.3=30\%\) of the students surveyed like football. And it is the number you would use to estimate that a randomly chosen student from this school likes football, whereas the raw frequency $\displaystyle 15$ could not be used that way - $\displaystyle 15$ out of $\displaystyle 50$ and $\displaystyle 15$ out of $\displaystyle 5000$ are very different situations.Answer: the blanks are "relative frequency" and \(\displaystyle \dfrac{15}{50}=\dfrac{3}{10}=0.3\) (that is, \(\displaystyle 30\%\)).

The test for "equally likely". Two or more outcomes are equally likely when there is a genuine reason - usually symmetry, or identical construction - for each to occur just as often as the others. Simply having two possible outcomes is not enough; that is the mistake this question is checking for.(i) The car starts, or does not start. NOT equally likely. There are two outcomes, but nothing makes them balanced. A car in good condition starts nearly every time, so "starts" is far more likely than "does not start"; a car with a dead battery is the other way round. The probabilities depend on the state of the car and cannot be found by counting outcomes.(ii) Tossing a fair coin once. Equally likely. The word fair means the coin is symmetric and unbiased, so \(\displaystyle S=\{H,T\}\) with \[P(H)=P(T)=\frac{1}{2} \](iii) Rolling a fair $\displaystyle 6$-sided die. Equally likely. A fair die is a cube with identical faces, so no face is favoured. \(\displaystyle S=\{1,2,3,4,5,6\}\) and \[P(1)=P(2)=\cdots=P(6)=\frac{1}{6} \](iv) A marble from a bag of $\displaystyle 3$ red and $\displaystyle 7$ blue. NOT equally likely - as far as the two colours go. Blue outnumbers red more than twice over: \[P(\text{red})=\frac{3}{10},\qquad P(\text{blue})=\frac{7}{10} \] The subtlety worth noticing: the ten marbles are equally likely to be drawn, since they are identical apart from colour. It is the two colour outcomes that are unequal, because $\displaystyle 7$ of the equally likely marbles produce "blue" and only $\displaystyle 3$ produce "red".(v) A baby is born - boy or girl. Treated as equally likely. In this chapter we take \(\displaystyle P(\text{boy})=P(\text{girl})=\tfrac12\). Being honest about the real world: records show slightly more boys are born than girls - roughly $\displaystyle 51$ boys per $\displaystyle 100$ births - so the two are very nearly, but not exactly, equally likely. It is a good example of a model that is close enough to be useful.Answers: equally likely - (ii) tossing a fair coin and (iii) rolling a fair die, by symmetry; (v) a boy or a girl, to a very good approximation. Not equally likely - (i) the car starting, which depends entirely on the car's condition, and (iv) red or blue, since $\displaystyle 3$ marbles give red but $\displaystyle 7$ give blue.

One method, five times over. In each part, write the sample space of equally likely outcomes, count the ones that make the event happen, and divide: \[P(E)=\frac{n(E)}{n(S)} \](i) Two coins, at least one head. Each coin can land \(\displaystyle H\) or \(\displaystyle T\), and the coins are separate, so there are \(\displaystyle 2\times 2=4\) outcomes: \[S=\{HH,\ HT,\ TH,\ TT\},\qquad n(S)=4 \] "At least one head" means one or two heads - so everything except \(\displaystyle TT\): \[E=\{HH,\ HT,\ TH\},\qquad n(E)=3,\qquad P=\frac{3}{4} \] A quicker route to the same number: the only way to fail is \(\displaystyle TT\), which has probability \(\displaystyle \tfrac14\), so \(\displaystyle P=1-\tfrac14=\tfrac34\ \checkmark\). (Note that \(\displaystyle HT\) and \(\displaystyle TH\) must be counted separately - the first coin showing heads is a different result from the second coin showing heads.)(ii) A card numbered $\displaystyle 1$ to $\displaystyle 10$, an even number. \[S=\{1,2,3,4,5,6,7,8,9,10\},\qquad n(S)=10 \] \[E=\{2,4,6,8,10\},\qquad n(E)=5,\qquad P=\frac{5}{10}=\frac{1}{2} \] The cards are identical, so each is equally likely to be drawn.(iii) A die, a number greater than 4. \[S=\{1,2,3,4,5,6\},\qquad n(S)=6 \] "Greater than $\displaystyle 4$" does not include $\displaystyle 4$ itself: \[E=\{5,6\},\qquad n(E)=2,\qquad P=\frac{2}{6}=\frac{1}{3} \](iv) $\displaystyle 3$ red, $\displaystyle 2$ blue, $\displaystyle 1$ green - not red. The bag holds \(\displaystyle 3+2+1=6\) balls, all equally likely to be picked, so take the $\displaystyle 6$ balls as the sample space. "Not red" means blue or green: \[n(\text{not red})=2+1=3,\qquad P=\frac{3}{6}=\frac{1}{2} \] Check with the complement: \(\displaystyle P(\text{red})=\tfrac36=\tfrac12\), and \(\displaystyle 1-\tfrac12=\tfrac12\ \checkmark\).(v) Three coins, exactly two heads. Each of the three coins has $\displaystyle 2$ possibilities, so \(\displaystyle n(S)=2\times 2\times 2=8\): \[S=\{HHH,\ HHT,\ HTH,\ HTT,\ THH,\ THT,\ TTH,\ TTT\} \] "Exactly two heads" means two heads and one tail - and the tail can be on any one of the three coins: \[E=\{HHT,\ HTH,\ THH\},\qquad n(E)=3,\qquad P=\frac{3}{8} \] Be careful with the word exactly: \(\displaystyle HHH\) has two heads among its three, but it is not in the event, because it has three heads and no tail.Answers: (i) \(\displaystyle \dfrac{3}{4}\); (ii) \(\displaystyle \dfrac{5}{10}=\dfrac{1}{2}\); (iii) \(\displaystyle \dfrac{2}{6}=\dfrac{1}{3}\); (iv) \(\displaystyle \dfrac{3}{6}=\dfrac{1}{2}\); (v) \(\displaystyle \dfrac{3}{8}\).

Three candies, one of each kind. The bag holds three distinct candies and one is taken at random, which means each candy is equally likely to be the one chosen. The sample space is \[S=\{\text{strawberry},\ \text{lemon},\ \text{mint}\},\qquad n(S)=3 \] Exactly one of these three is the strawberry, so \[E=\{\text{strawberry}\},\qquad n(E)=1 \] \[P(\text{strawberry})=\frac{n(E)}{n(S)}=\frac{1}{3} \]Here the three outcomes really are equally likely, because there is exactly one candy of each flavour - which is why we may divide by 3. Had the bag held $\displaystyle 2$ strawberry candies and $\displaystyle 1$ each of the others, the flavours would no longer be equally likely and the answer would be \(\displaystyle \tfrac{2}{4}\), not \(\displaystyle \tfrac13\).Check. The three flavours have probabilities \(\displaystyle \tfrac13,\tfrac13,\tfrac13\), and \(\displaystyle \tfrac13+\tfrac13+\tfrac13=1\), as the probabilities of all the outcomes must.Answer: \(\displaystyle P(\text{strawberry})=\dfrac{1}{3}\).

Every shirt with every pair of pants. An outfit is one choice of shirt together with one choice of pants. The two choices are made independently, so each of the $\displaystyle 2$ shirts can be worn with each of the $\displaystyle 3$ pants. Building the table with the shirts as rows and the pants as columns makes every combination appear exactly once - and makes the count obvious, since the table has \(\displaystyle 2\) rows \(\displaystyle \times\ 3\) columns \(\displaystyle =6\) cells.

	Jeans	Khakis	Shorts
Red shirt	Red shirt \(\displaystyle +\) jeans	Red shirt \(\displaystyle +\) khakis	Red shirt \(\displaystyle +\) shorts
Blue shirt	Blue shirt \(\displaystyle +\) jeans	Blue shirt \(\displaystyle +\) khakis	Blue shirt \(\displaystyle +\) shorts

Listed out, the six outfits are:1. Red shirt with jeans 2. Red shirt with khakis 3. Red shirt with shorts 4. Blue shirt with jeans 5. Blue shirt with khakis 6. Blue shirt with shortsWhy \(\displaystyle 2\times 3\). Fix the red shirt: there are $\displaystyle 3$ outfits, one for each pair of pants. Fix the blue shirt: another 3. Nothing else is possible and nothing is repeated, so \(\displaystyle 3+3=6\) - which is what the multiplication \(\displaystyle 2\times 3=6\) is a shortcut for. This is the same counting rule that gave $\displaystyle 6$ snack-and-drink combinations at the village fair, and it is worth remembering: when one choice of \(\displaystyle m\) kinds is combined with an independent choice of \(\displaystyle n\) kinds, there are \(\displaystyle m\times n\) combinations.Answer: there are \(\displaystyle 2\times 3=6\) possible outfits - red with jeans, red with khakis, red with shorts, blue with jeans, blue with khakis, and blue with shorts - as set out in the table above.

Experimental probability, because a tyre is not a fair die.Nothing here can be argued from symmetry. All we have is what happened in $\displaystyle 1000$ recorded cases, so every probability is\[P(\text{event})=\frac{\text{number of cases in which it happened}}{\text{total number of cases}} \]First check the table accounts for every case:\[20+210+325+445=1000 \]That is exactly the number recorded, so nothing is missing and we may divide by $\displaystyle 1000$ throughout.(i) Lasts less than $\displaystyle 4000$ kmThis is one row of the table, read straight off:\[P=\frac{20}{1000}=\frac{1}{50}=0.02 \](ii) Lasts between $\displaystyle 4000$ and $\displaystyle 14000$ kmNo single row covers this. Two rows do — "$\displaystyle 4001$ to $\displaystyle 9000$" and "$\displaystyle 9001$ to $\displaystyle 14000$" — and a tyre counted in one cannot also be counted in the other, so the counts simply add:\[P=\frac{210+325}{1000}=\frac{535}{1000}=\frac{107}{200}=0.535 \](iii) Lasts more than $\displaystyle 14000$ kmAgain one row:\[P=\frac{445}{1000}=\frac{89}{200}=0.445 \]A check worth doing. The three cases asked about are "under $\displaystyle 4000$", "$\displaystyle 4000$ to $\displaystyle 14000$" and "over $\displaystyle 14000$" — no tyre falls in two of them, and every tyre falls in one. So the three probabilities must total $\displaystyle 1$:\[0.02+0.535+0.445=1 \]They do, which means no row was dropped or double-counted.Answer: (i) \(\displaystyle 0.02\) (ii) \(\displaystyle 0.535\) (iii) \(\displaystyle 0.445\)

Counting equally likely cards.Write the word out letter by letter, because that is what is on the cards:\[\text{P},\ \text{E},\ \text{A},\ \text{C},\ \text{E} \]There are $\displaystyle 5$ cards, so the sample space has $\displaystyle 5$ outcomes, and since Leela draws without looking each card is equally likely. The one thing to be careful about is that E appears twice — those are two different cards that happen to carry the same letter, and both of them count.(i) A P, an E or a C.Favourable cards: P ($\displaystyle 1$ card), E ($\displaystyle 2$ cards), C ($\displaystyle 1$ card), so \(\displaystyle 1 + 2 + 1 = 4\) favourable cards out of 5.\[P(\text{P, E or C}) = \frac{4}{5} \]A second way to see it: the only card that is not a P, E or C is the A, so \(\displaystyle P = 1 - \frac{1}{5} = \frac{4}{5}\). The two methods agree.(ii) Not an E.Two of the five cards are E, so \(\displaystyle 5 - 2 = 3\) cards are not E:\[P(\text{not E}) = \frac{3}{5} \]Check with the complement rule: \(\displaystyle P(\text{E}) = \frac{2}{5}\), and \(\displaystyle \frac{2}{5} + \frac{3}{5} = 1\), as it must be.Answer: (i) \(\displaystyle \frac{4}{5}\) (ii) \(\displaystyle \frac{3}{5}\)

Equally likely outcomes.The figure shows the spinner, but everything we need is already stated in words: the arrow stops on one of \[1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8 \] and these $\displaystyle 8$ outcomes are equally likely. So the sample space has $\displaystyle 8$ outcomes and\[P(\text{event}) = \frac{\text{number of favourable numbers}}{8} \](i) Pointing at 8. Just one of the eight numbers is an 8. \[P = \frac{1}{8} \](ii) An odd number. The odd numbers are \(\displaystyle 1, 3, 5, 7\) — that is $\displaystyle 4$ of them. \[P = \frac{4}{8} = \frac{1}{2} \](iii) A number greater than 2. "Greater than $\displaystyle 2$" does not include $\displaystyle 2$ itself, so the favourable numbers are \(\displaystyle 3, 4, 5, 6, 7, 8\) — $\displaystyle 6$ of them. \[P = \frac{6}{8} = \frac{3}{4} \](iv) A number less than 9. Every one of \(\displaystyle 1, 2, \ldots, 8\) is less than $\displaystyle 9$, so all $\displaystyle 8$ outcomes are favourable. This is a sure event. \[P = \frac{8}{8} = 1 \](v) A multiple of 3. Among $\displaystyle 1$ to $\displaystyle 8$ the multiples of $\displaystyle 3$ are \(\displaystyle 3\) and \(\displaystyle 6\) — note that $\displaystyle 9$ is a multiple of $\displaystyle 3$ but is not on the spinner. \[P = \frac{2}{8} = \frac{1}{4} \]Answer: (i) \(\displaystyle \frac{1}{8}\) (ii) \(\displaystyle \frac{1}{2}\) (iii) \(\displaystyle \frac{3}{4}\) (iv) \(\displaystyle 1\) (v) \(\displaystyle \frac{1}{4}\)

A two-stage tree, with the second stage depending on the first.The basket holds \(\displaystyle 4 + 5 = 9\) balls. The first ball is laid aside, so when the second ball is drawn only $\displaystyle 8$ balls remain — and one colour is short by one, depending on what came out first. That dependence is exactly what a tree diagram is for.The tree in words. Start at a single dot. Two branches leave it, for the first draw: Now grow two more branches out of each of those, for the second draw: That gives $\displaystyle 4$ complete paths. To get the probability of a path, multiply the probabilities along it:

First ball	Second ball	Multiply along the path	Probability
Red \(\displaystyle \left(\frac{4}{9}\right)\)	Red \(\displaystyle \left(\frac{3}{8}\right)\)	\(\displaystyle \frac{4}{9} \times \frac{3}{8} = \frac{12}{72}\)	\(\displaystyle \frac{1}{6}\)
Red \(\displaystyle \left(\frac{4}{9}\right)\)	Blue \(\displaystyle \left(\frac{5}{8}\right)\)	\(\displaystyle \frac{4}{9} \times \frac{5}{8} = \frac{20}{72}\)	\(\displaystyle \frac{5}{18}\)
Blue \(\displaystyle \left(\frac{5}{9}\right)\)	Red \(\displaystyle \left(\frac{4}{8}\right)\)	\(\displaystyle \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\)	\(\displaystyle \frac{5}{18}\)
Blue \(\displaystyle \left(\frac{5}{9}\right)\)	Blue \(\displaystyle \left(\frac{4}{8}\right)\)	\(\displaystyle \frac{5}{9} \times \frac{4}{8} = \frac{20}{72}\)	\(\displaystyle \frac{5}{18}\)

Check: \(\displaystyle \frac{12}{72} + \frac{20}{72} + \frac{20}{72} + \frac{20}{72} = \frac{72}{72} = 1\). The four paths are the only things that can happen, so their probabilities must total 1.(i) Red first, then blue. This is the second path: \[P(\text{red, then blue}) = \frac{4}{9} \times \frac{5}{8} = \frac{20}{72} = \frac{5}{18} \](ii) Two blue balls. This is the fourth path: \[P(\text{blue, then blue}) = \frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \frac{5}{18} \]A second method, as a check. Imagine the $\displaystyle 9$ balls are numbered so we can tell them apart. An outcome is an ordered pair (first ball, second ball) with the two balls different, and there are \(\displaystyle 9 \times 8 = 72\) such equally likely pairs. Red-then-blue can happen in \(\displaystyle 4 \times 5 = 20\) ways, giving \(\displaystyle \frac{20}{72} = \frac{5}{18}\); blue-then-blue in \(\displaystyle 5 \times 4 = 20\) ways, giving \(\displaystyle \frac{20}{72} = \frac{5}{18}\). Both match the tree.(It is a coincidence worth noticing, not a rule, that these two answers came out equal: \(\displaystyle 4 \times 5\) and \(\displaystyle 5 \times 4\) are the same number.)Answer: (i) \(\displaystyle \frac{5}{18}\) (ii) \(\displaystyle \frac{5}{18}\)

Impossible events and sure events.Throwing a pair of $\displaystyle 6$-sided dice, each die shows one of \(\displaystyle 1, 2, 3, 4, 5, 6\), so there are \(\displaystyle 6 \times 6 = 36\) equally likely outcomes, written as ordered pairs like \(\displaystyle (3, 5)\). The two numbers add up to at least \(\displaystyle 1 + 1 = 2\) and at most \(\displaystyle 6 + 6 = 12\).An event with probability 0. Take the event "the sum of the two numbers is $\displaystyle 13$". No pair of numbers from $\displaystyle 1$ to $\displaystyle 6$ can add to $\displaystyle 13$, so not one of the $\displaystyle 36$ outcomes is favourable: \[P(\text{sum is } 13) = \frac{0}{36} = 0 \] An event of probability $\displaystyle 0$ is called an impossible event. Other correct answers: "the sum is $\displaystyle 1$", "a die shows $\displaystyle 7$", "both dice show $\displaystyle 0$".An event with probability 1. Take the event "the sum is one of \(\displaystyle 2, 3, 4, \ldots, 12\)". Every one of the $\displaystyle 36$ outcomes is favourable: \[P(\text{sum is between } 2 \text{ and } 12) = \frac{36}{36} = 1 \] An event of probability $\displaystyle 1$ is called a sure (or certain) event. Other correct answers: "each die shows a whole number from $\displaystyle 1$ to $\displaystyle 6$", "the sum is at least $\displaystyle 2$".One point to be careful about. The question asks for an outcome of probability 1. For this experiment a single outcome is one pair, such as \(\displaystyle (3, 5)\), and each of the $\displaystyle 36$ pairs has probability \(\displaystyle \frac{1}{36}\) — so no single outcome can have probability 1. What has probability $\displaystyle 1$ is an event that collects together all $\displaystyle 36$ outcomes, like the one above. (A single outcome could have probability $\displaystyle 1$ only in an experiment with just one possible result.)Many different answers are equally valid here; any impossible event and any certain event will do.Answer: for example, \(\displaystyle P(\text{"the sum is }13\text{"}) = 0\) and \(\displaystyle P(\text{"the sum is one of }2,3,\ldots,12\text{"}) = 1\). Other impossible and certain events are equally correct.

(i) Two dice, sum a prime greater than 5.List the sample space first. Rolling two dice, an outcome is an ordered pair \(\displaystyle (\text{first die}, \text{second die})\), and there are \(\displaystyle 6 \times 6 = 36\) equally likely outcomes. The possible sums run from $\displaystyle 2$ to 12.Which of those sums are prime? \(\displaystyle 2, 3, 5, 7, 11\). Of these, the ones greater than $\displaystyle 5$ are \(\displaystyle 7\) and \(\displaystyle 11\) only. ($\displaystyle 9$ is not prime, \(\displaystyle 9 = 3 \times 3\); neither is 12.)Sum \(\displaystyle = 7\): \(\displaystyle (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\) — $\displaystyle 6$ outcomes. Sum \(\displaystyle = 11\): \(\displaystyle (5,6), (6,5)\) — $\displaystyle 2$ outcomes.These lists do not overlap, so there are \(\displaystyle 6 + 2 = 8\) favourable outcomes. \[P = \frac{8}{36} = \frac{2}{9} \](ii) Two balls of different colours.The bag has \(\displaystyle 4 + 3 + 2 = 9\) balls. Two are drawn without replacement, and the order they come out in does not matter here, so an outcome is a pair of balls. The number of pairs is \[\frac{9 \times 8}{2} = 36 \] (\(\displaystyle 9 \times 8\) counts ordered picks; we divide by $\displaystyle 2$ because each pair gets counted twice, once in each order.)It is quicker to count the opposite event, "both the same colour": \[\text{both red: } \frac{4 \times 3}{2} = 6, \qquad \text{both green: } \frac{3 \times 2}{2} = 3, \qquad \text{both blue: } \frac{2 \times 1}{2} = 1 \] so \(\displaystyle 6 + 3 + 1 = 10\) same-colour pairs, and \(\displaystyle 36 - 10 = 26\) different-colour pairs. \[P(\text{different colours}) = \frac{26}{36} = \frac{13}{18} \] Direct check: red-green \(\displaystyle 4 \times 3 = 12\), red-blue \(\displaystyle 4 \times 2 = 8\), green-blue \(\displaystyle 3 \times 2 = 6\); \(\displaystyle 12 + 8 + 6 = 26\). It agrees.(iii) Three coins: first is a head, and exactly two heads in all.Sample space ($\displaystyle 8$ equally likely outcomes): \[\text{HHH},\ \text{HHT},\ \text{HTH},\ \text{HTT},\ \text{THH},\ \text{THT},\ \text{TTH},\ \text{TTT} \] We need both conditions at once: the first letter is H, and there are exactly two H's altogether. Going through the list, HHT and HTH qualify. (THH has exactly two heads but starts with a tail; HHH starts with a head but has three heads.) \[P = \frac{2}{8} = \frac{1}{4} \](iv) A four-digit number from $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$ with no repetition, even.Every such number uses each digit exactly once, so the sample space is all the arrangements of \(\displaystyle 1, 2, 3, 4\): \[4 \times 3 \times 2 \times 1 = 24 \text{ numbers} \] A number is even exactly when its last digit is even, so the last digit must be $\displaystyle 2$ or $\displaystyle 4$ — $\displaystyle 2$ choices. The other three digits can then be arranged in the remaining three places in \(\displaystyle 3 \times 2 \times 1 = 6\) ways, giving \(\displaystyle 2 \times 6 = 12\) even numbers. \[P(\text{even}) = \frac{12}{24} = \frac{1}{2} \] This is believable: of the four digits, two are even and two are odd, so half the arrangements should end in an even digit.(v) Guessing $\displaystyle 3$ multiple-choice questions, exactly $\displaystyle 2$ right.An outcome is the student's set of three answers. Each question has $\displaystyle 4$ options, so \[4 \times 4 \times 4 = 64 \text{ equally likely answer sheets} \] For exactly $\displaystyle 2$ correct, first choose which question is the wrong one — $\displaystyle 3$ ways. The other two must be answered with the single correct option ($\displaystyle 1$ way each), and the wrong one must be answered with one of the $\displaystyle 3$ wrong options. \[3 \times 1 \times 1 \times 3 = 9 \text{ favourable answer sheets} \] \[P(\text{exactly 2 correct}) = \frac{9}{64} \] (A common slip is to forget that the "wrong" question can be wrong in $\displaystyle 3$ different ways and to write \(\displaystyle \frac{3}{64}\).)Answer: (i) \(\displaystyle \frac{2}{9}\) (ii) \(\displaystyle \frac{13}{18}\) (iii) \(\displaystyle \frac{1}{4}\) (iv) \(\displaystyle \frac{1}{2}\) (v) \(\displaystyle \frac{9}{64}\)

Growing a tree, with and without replacement.The box holds the balls \(\displaystyle 1, 2, 3, 4\). In both experiments the first draw is the same: a single starting dot with $\displaystyle 4$ branches, labelled $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, 4. The two experiments differ only in what grows out of each of those branches.(i) The ball is returned before the second draw.The box is back to its full contents, so every first branch splits into $\displaystyle 4$ second branches, labelled $\displaystyle 1$, $\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$ — including a repeat of the number just drawn. Writing an outcome as (first, second), the leaves of the tree are:

First ball (row) / Second ball (column)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$
$\displaystyle 1$	$\displaystyle (1, 1)$	$\displaystyle (1, 2)$	$\displaystyle (1, 3)$	$\displaystyle (1, 4)$
$\displaystyle 2$	$\displaystyle (2, 1)$	$\displaystyle (2, 2)$	$\displaystyle (2, 3)$	$\displaystyle (2, 4)$
$\displaystyle 3$	$\displaystyle (3, 1)$	$\displaystyle (3, 2)$	$\displaystyle (3, 3)$	$\displaystyle (3, 4)$
$\displaystyle 4$	$\displaystyle (4, 1)$	$\displaystyle (4, 2)$	$\displaystyle (4, 3)$	$\displaystyle (4, 4)$

(Each row is one first-stage branch and its four second-stage branches.)(ii) The first ball is not replaced.Only $\displaystyle 3$ balls are left for the second draw, and the number already drawn cannot appear again. So every first branch splits into just $\displaystyle 3$ second branches. The four repeats \(\displaystyle (1,1), (2,2), (3,3), (4,4)\) are missing; a dash marks where they would have been:

First ball (row) / Second ball (column)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$
$\displaystyle 1$	---	$\displaystyle (1, 2)$	$\displaystyle (1, 3)$	$\displaystyle (1, 4)$
$\displaystyle 2$	$\displaystyle (2, 1)$	---	$\displaystyle (2, 3)$	$\displaystyle (2, 4)$
$\displaystyle 3$	$\displaystyle (3, 1)$	$\displaystyle (3, 2)$	---	$\displaystyle (3, 4)$
$\displaystyle 4$	$\displaystyle (4, 1)$	$\displaystyle (4, 2)$	$\displaystyle (4, 3)$	---

(iii) Sizes of the two sample spaces.With replacement: $\displaystyle 4$ first branches, each splitting into $\displaystyle 4$, so \[4 \times 4 = 16 \text{ outcomes.} \] Without replacement: $\displaystyle 4$ first branches, each splitting into $\displaystyle 3$, so \[4 \times 3 = 12 \text{ outcomes.} \] The second is smaller by exactly the $\displaystyle 4$ repeats \(\displaystyle (1,1), (2,2), (3,3), (4,4)\), which is a good way to remember why: \(\displaystyle 16 - 4 = 12\).Answer: (i) $\displaystyle 16$ outcomes, \(\displaystyle (1,1)\) through \(\displaystyle (4,4)\); (ii) the same list with the $\displaystyle 4$ repeats removed; (iii) the sample spaces have sizes $\displaystyle 16$ and 12.

Pair every coin result with every card.The two things happen at the same time and neither affects the other, so a single outcome has to record both: what the coin did and which card was drawn. Write it as an ordered pair, coin first.The coin gives $\displaystyle 2$ results, H or T. The card gives $\displaystyle 6$ results, $\displaystyle 1$ to 6. Each of the $\displaystyle 2$ coin results can go with each of the $\displaystyle 6$ cards, so the sample space has \[2 \times 6 = 12 \text{ outcomes.} \]\[S = \{(\text{H},1),\ (\text{H},2),\ (\text{H},3),\ (\text{H},4),\ (\text{H},5),\ (\text{H},6),\ (\text{T},1),\ (\text{T},2),\ (\text{T},3),\ (\text{T},4),\ (\text{T},5),\ (\text{T},6)\} \]Laying it out as a grid makes the \(\displaystyle 2 \times 6\) clear, and also makes it easy to check that nothing has been left out or written twice:

	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$
H	(H, $\displaystyle 1$)	(H, $\displaystyle 2$)	(H, $\displaystyle 3$)	(H, $\displaystyle 4$)	(H, $\displaystyle 5$)	(H, $\displaystyle 6$)
T	(T, $\displaystyle 1$)	(T, $\displaystyle 2$)	(T, $\displaystyle 3$)	(T, $\displaystyle 4$)	(T, $\displaystyle 5$)	(T, $\displaystyle 6$)

If the coin is fair and the card is drawn at random, all $\displaystyle 12$ outcomes are equally likely, each with probability \(\displaystyle \frac{1}{12}\).Answer: \(\displaystyle S = \{(\text{H},1), (\text{H},2), (\text{H},3), (\text{H},4), (\text{H},5), (\text{H},6), (\text{T},1), (\text{T},2), (\text{T},3), (\text{T},4), (\text{T},5), (\text{T},6)\}\), which has $\displaystyle 12$ elements.

What a list must do to be a sample space.A sample space is the list of results of the experiment, and it has to satisfy two conditions at once:1. Nothing missing — every time the experiment is performed, the result is somewhere on the list; 2. Nothing extra — everything on the list is a result that really can happen.Here the experiment is: toss three coins and record the number of heads. So an outcome is a number, not a string like HTH. The fewest heads possible is $\displaystyle 0$ (the toss TTT) and the most is $\displaystyle 3$ (the toss HHH), and $\displaystyle 1$ and $\displaystyle 2$ clearly happen too. So the possible records are exactly \(\displaystyle 0, 1, 2, 3\).That is list (iv), \(\displaystyle \{0, 1, 2, 3\}\).Why the others fail: One thing that is not required. The four outcomes \(\displaystyle 0, 1, 2, 3\) are not equally likely — from the eight equally likely tosses HHH, HHT, HTH, HTT, THH, THT, TTH, TTT the probabilities work out as \(\displaystyle \frac{1}{8}, \frac{3}{8}, \frac{3}{8}, \frac{1}{8}\). That does not stop \(\displaystyle \{0, 1, 2, 3\}\) being the sample space; a sample space only has to be complete and correct, not evenly balanced.Answer: (iv) \(\displaystyle \{0, 1, 2, 3\}\). (i) and (ii) are incomplete ($\displaystyle 0$ and $\displaystyle 3$ respectively are missing), and (iii) includes $\displaystyle 4$, which is impossible with three coins.

Geometric probability: when every point is equally likely, probability is a ratio of AREAS, not of counts.The dye can land anywhere on the rectangle, and no spot is favoured over another. So\[P(\text{lands in the circle})=\frac{\text{area of the circle}}{\text{area of the rectangle}} \]Area of the rectangle. From the figure its sides are $\displaystyle 3$ m and $\displaystyle 2$ m:\[\text{area}=3\times 2=6\ \text{m}^{2} \]Area of the circle. The question gives the diameter as $\displaystyle 1$ m, so the radius is half of that — this is the step to be careful with:\[r=\frac{1}{2}\ \text{m}, \qquad \text{area}=\pi r^{2}=\pi\left(\frac{1}{2}\right)^{2}=\frac{\pi}{4}\ \text{m}^{2} \]Divide:\[P=\frac{\pi/4}{6}=\frac{\pi}{24} \]Taking \(\displaystyle \pi=\frac{22}{7}\):\[P=\frac{22}{7\times 24}=\frac{22}{168}=\frac{11}{84}\approx 0.13 \]So a little over one drop in eight lands inside the circle. That is believable from the picture — the circle covers a modest part of the rectangle, nowhere near half of it.Answer: \(\displaystyle P=\dfrac{\pi}{24}=\dfrac{11}{84}\approx 0.13\)