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

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) \(\displaystyle \frac{5}{18}\)