Class 9 Maths Ganita Manjari Chapter 7 Solutions — The Mathematics of Maybe: Introduction to Probability

Sample classifications, with ordinary school-day assumptions: (i) Certain, probability 1: in the weekly calendar Monday follows Sunday. (ii) Extremely unlikely, near 0 under ordinary Mumbai weather conditions; the classroom exercise may treat this as practically impossible. No measured numerical probability is provided. (iii) Extremely unlikely in an ordinary classroom; call it impossible only if the setting makes entry physically impossible. (iv) More likely than not if tomorrow is a school day, you attend and usually meet friends; it is not certain.

The last three depend on context. “Very unlikely” is not automatically the same as mathematical probability 0. Do not invent precise probabilities from these descriptions.

(i) Within the 30-sweet sample, $P\left(\text{green}\right)=\frac{8}{30}=\frac{4}{15}\approx 0.267$. (ii) The sample yellow proportion is $\frac{7}{30}$, so the estimated number in the 600-sweet bag is $\left(\frac{7}{30}\right)\times 600=140$ sweets. This extrapolation assumes the mixed random sample represents the bag; 140 is an estimate, not a guaranteed count.

(i) Within the sample, $P\left(\text{Arts}\right)=\frac{11}{40}=0.275$. (ii) Estimated school-wide Sports preference=$\left(\frac{9}{40}\right)\times 800=180$ students. This is an estimate based on a representative random sample, not an exact school census.

Record your own 20 results. If h are heads, then (i) heads=h, (ii) tails=20−h, and (iii) experimental $P\left(\text{heads}\right)=\frac{h}{20}$. Illustrative example only: 12 heads and 8 tails give $\frac{12}{20}=0.6$. These are not claimed experimental observations.

(iv) Assuming a fair coin and independent tosses, $$P\left(\text{tails on the next toss}\right)=\frac{1}{2}\text{,}$$ regardless of the previous count.

Let b,t,s be the observed counts for bottom, top and side, with $b+t+s=100$. Then the experimental probabilities are $\frac{b}{100}$,$\frac{t}{100}$,$\frac{s}{100}$; they must sum to 1.

Replace these example values with your own data. The three positions need not be equally likely: the cup is not symmetric in those orientations. Having three outcomes does not justify giving each probability $\frac{1}{3}$.

(i) Experimental probability=$\frac{3}{12}=\frac{1}{4}$. (ii) For a fair die, theoretical probability=$\frac{1}{6}$. (iii) The observed relative frequency fluctuates by chance, especially with 12 trials. With 60,600 or 6000 independent rolls we expect it to be more stable around $\frac{1}{6}$, but no particular finite set is guaranteed to be closer, or to contain exactly 10,100 or 1000 threes.

(i) $$S=\{\left(1,H\right),\left(1,T\right),\left(2,H\right),\left(2,T\right),\left(3,H\right),\left(3,T\right),\left(4,H\right),\left(4,T\right),\left(5,H\right),\left(5,T\right),\left(6,H\right),\left(6,T\right)\}\text{.}$$ There are 12 outcomes, equally likely if the coin and die are fair and independent.

(ii) Interpreting “between −5 and +5” inclusively: $$S=\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}\text{.}$$ If “strictly between” is intended, omit −5 and 5; the textbook does not explicitly specify endpoints.

(iii) By colour, S={Green,Red}, but these outcomes are not equally likely: $P\left(\text{Green}\right)=\frac{5}{12}$ and $P\left(\text{Red}\right)=\frac{7}{12}$. An equally likely individual-ball space is $$\{G_{1},G_{2},G_{3},G_{4},G_{5},R_{1},R_{2},R_{3},R_{4},R_{5},R_{6},R_{7}\}\text{.}$$

Label the oranges $O_{1}$,$O_{2}$ and the apple A; write B for banana and M for mango. The equally likely sample space is $$\{\left(A,B\right),\left(A,M\right),\left(O_{1},B\right),\left(O_{1},M\right),\left(O_{2},B\right),\left(O_{2},M\right)\}$$. Each pair has probability $\frac{1}{6}$ under independent uniform selections. The colour/type-level list has four pairs, but they do not all have equal probabilities.

$$\begin{aligned}P\left(\text{apple and banana}\right)&=\left(\frac{1}{3}\right)\left(\frac{1}{2}\right)\\&=\frac{1}{6}\text{.}\end{aligned}$$

Let R=red, K=black, G=green. The colour sample space is {RR,RK,RG,KR,KK,KG,GR,GK,GG}. With replacement, each second draw has the original probabilities $\frac{3}{9}$,$\frac{4}{9}$,$\frac{2}{9}$. These nine colour pairs are not equally likely.

$$\begin{aligned}P\left(\text{same colour}\right)&=\left(\frac{3}{9}\right)^{2}+\left(\frac{4}{9}\right)^{2}+\left(\frac{2}{9}\right)^{2}\\&=\frac{9+16+4}{81}\\&=\frac{29}{81}\\&\approx 0.358\text{.}\end{aligned}$$

(i) Not necessarily: a car’s starting probability depends on its condition. (ii) Yes, H and T each have probability $\frac{1}{2}$ for a fair coin. (iii) Yes, each face has probability $\frac{1}{6}$ for a fair die. (iv) Individual marbles are equally likely under uniform selection, but colours are not: $P\left(\text{red}\right)=\frac{3}{10}$ and $P\left(\text{blue}\right)=\frac{7}{10}$.

(v) For this textbook’s simplified two-outcome model, boy/girl is commonly treated as approximately equally likely. Exact equality is an assumption of that model, not a consequence of merely listing two possibilities.

Assume fair independent coins/dice and uniform selections.

(i) $S=\{\mathrm{HH},\mathrm{HT},\mathrm{TH},\mathrm{TT}\}$; $E=\{\mathrm{HH},\mathrm{HT},\mathrm{TH}\}$, so $P\left(E\right)=\frac{3}{4}$.

(ii) $S=\{1,2,3,4,5,6,7,8,9,10\}$; $E=\{2,4,6,8,10\}$, so $P\left(E\right)=\frac{1}{2}$.

(iii) S={1,2,3,4,5,6}; $E=\{5,6\}$, so $P\left(E\right)=\frac{1}{3}$.

(iv) Label the balls: $$S=\{R_{1},R_{2},R_{3},B_{1},B_{2},G_{1}\}\text{;}$$ $E=\{B_{1},B_{2},G_{1}\}$, so $P\left(\text{not red}\right)=\frac{1}{2}$.

(v) $$S=\{\mathrm{HHH},\mathrm{HHT},\mathrm{HTH},\mathrm{HTT},\mathrm{THH},\mathrm{THT},\mathrm{TTH},\mathrm{TTT}\}\text{;}$$ $E=\{\mathrm{HHT},\mathrm{HTH},\mathrm{THH}\}$, so $P\left(\text{exactly two heads}\right)=\frac{3}{8}$.

There are $2\times 3=6$ outfits. The table lists combinations; no equal-probability assumption about the child’s preferences is needed.

(i) $\frac{20}{1000}=0.02$. (ii) Add the two middle classes: $\frac{210+325}{1000}=0.535$. (iii) $\frac{445}{1000}=0.445$. The three proportions sum to 1.

Class-boundary caveat: the printed intervals omit exactly 4000 km (and have gaps if distances are continuous). The answer 0.535 follows the intended grouping of the two middle classes. For a precisely inclusive interval, the missing boundary convention would need clarification. These frequencies describe the recorded sample; predicting a new tyre requires a representative-data assumption.

Distinguish the two E cards: $S=\{P,E_{1},A,C,E_{2}\}$, with five equally likely cards. (i) Separately, $P\left(P\right)=\frac{1}{5}$, $P\left(E\right)=\frac{2}{5}$, $P\left(C\right)=\frac{1}{5}$. If “P, E or C” means their combined event, add the four favourable cards: $\frac{4}{5}$. (ii) Not E consists of {P,A,C}, giving $\frac{3}{5}$.

With no replacement, the second denominator is 8 and its colour counts depend on the first draw. (i) $$\begin{aligned}P\left(\text{red then blue}\right)&=\left(\frac{4}{9}\right)\left(\frac{5}{8}\right)\\&=\frac{5}{18}\text{.}\end{aligned}$$ (ii) $P\left(\text{two blue}\right)=\left(\frac{5}{9}\right)\left(\frac{4}{8}\right)=\frac{5}{18}$.

Sample impossible event: the sum is 1, so its probability is0. Sample certain event: the sum is one of 2,3,…,12, so its probability is1.

The textbook asks for an “outcome” with probability 1. Strictly, no individual ordered outcome of two fair independent dice has probability 1: each has probability $\frac{1}{36}$. The intended answer is a certain event, containing the whole sample space.

(i) S={(i,j): i,j∈{1,2,3,4,5,6}}, with 36 ordered outcomes. Favourable sums are 7 and 11: {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1),(5,6),(6,5)}. Thus $P=\frac{8}{36}=\frac{2}{9}$.

(ii) Label the nine balls $$U=\{R_{1},R_{2},R_{3},R_{4},G_{1},G_{2},G_{3},B_{1},B_{2}\}\text{.}$$ An equally likely sample space is S={(u,v):u,v∈U,u≠v}, with $9\times 8=72$ ordered outcomes. Different-colour outcomes number $2\left(4\times 3+4\times 2+3\times 2\right)=52$, so $P=\frac{52}{72}=\frac{13}{18}$. The factor2 counts both orders.

(iii) S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}. Favourable outcomes={HHT,HTH}, so $P=\frac{2}{8}=\frac{1}{4}$. “First” refers to the first labelled coin.

(iv) $$S=\{1234,1243,1324,1342,1423,1432,2134,2143,2314,2341,2413,2431,3124,3142,3214,3241,3412,3421,4123,4132,4213,4231,4312,4321\}\text{.}$$ Under uniform arrangement, even numbers end in 2 or 4; each choice leaves $3\times 2\times 1$ arrangements. Thus $P=\frac{12}{24}=\frac{1}{2}$.

(v) S={(x₁,x₂,x₃): each xᵢ∈{A,B,C,D}}, with $4^{3}=64$ equally likely answer triples under independent uniform guesses. Choose which of the 3 questions is wrong; it has 3 wrong options, while the other two answers are fixed. Thus 9 favourable triples and $P=\frac{9}{64}$.

(i) With replacement: S={(i,j):i,j∈{1,2,3,4}}, explicitly {11,12,13,14,21,22,23,24,31,32,33,34,41,42,43,44}; ij denotes first i then j. There are 16 equally likely outcomes.

(ii) Without replacement: $$S=\{12,13,14,21,23,24,31,32,34,41,42,43\}\text{,}$$ so 12 outcomes. The diagonal outcomes 11,22,33,44 are excluded.

If the recorded quantity is the number of heads, the sample space is option(iv), {0,1,2,3}. Option(i) omits 0, (ii) omits 3, and (iii) includes impossible 4.

These four counts are not equally likely. For three fair independent coins their probabilities are respectively $\frac{1}{8}$,$\frac{3}{8}$,$\frac{3}{8}$,$\frac{1}{8}$. This differs from the eight equally likely ordered H/T outcomes.

Assume the landing point is uniformly distributed over the rectangle and the entire circle lies inside it. Radius=$\frac{1}{2}m$, circle area=$\frac{\pi }{4}m^{2}$, rectangle area=$3\times 2=6m^{2}$. Therefore $P=\frac{\pi }{24}\approx 0.131$ (about 13.1%). “At random” must mean uniform by area for this calculation.