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

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: \(\displaystyle P=\dfrac{\pi}{24}=\dfrac{11}{84}\approx 0.13\)