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 \(\displaystyle \dfrac{1}{6}\).