**1** **In a cricket math, a bats woman hits a boundary $6$ times out of $30$ balls she plays. Find the probability that she did not hit a boundary.**

Number of times the bats woman hits a boundary = $6$$\\$ Total number of balls played = $30$$\\$ $\therefore $ Number of times that the bats woman does not hit a boundary $= 30 - 6 = 24$$\\$ $P (\text{she does not hit a boundary}) =$$\\$ $\dfrac{\text{Number of times when she does not hit boundary }}{\text{Total number of balls played }}$$\\$ $=\dfrac{24}{30}=\dfrac{4}{5}$

**2** **$1500$ families with $2$ children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having$\\$ $\begin{array}{|c|c|}\hline \text{Number of girls in a family} & 2 & 1 & 0 \\ \hline \text{Number of families} & 475 & 814 & 211 \\ \hline \end{array}$ $\\$ (i)$2$ girls$\\$ (ii)$1$ girl$\\$ (iii)No girl$\\$ Also check whether the sum of these probabilities is $1$.**

Total number of families $= 475 + 814 + 211 = 1500$$\\$ (i) Number of families having $2$ girls = $475$$\\$ $P_1$ (a randomly chosen family has $2$ girls) = $\\$ $\dfrac{\text{Number of families having 2 girls }}{\text{Total number of families }}$$\\$ $\dfrac{475}{1500}=\dfrac{19}{60}$$\\$ (ii) Number of families having $1$ girl = $814$$\\$ $P_2$ (a randomly chosen family has $1$ girl) =$\\$ $\dfrac{\text{ Number of families having 1 girl}}{\text{ Total number of families}}$$\\$ $=\dfrac{814}{1500}=\dfrac{407}{750}$$\\$ (iii) Number of families having no girl = $211$$\\$ $P_3$ (a randomly chosen family has no girl) =$\\$ $\dfrac{\text{Number of families having no girl } }{\text{ Total number of families }}$$\\$ $=\dfrac{211}{1500}$$\\$ Sum of all these probabilities $=\dfrac{19}{60}+\dfrac{407}{750}+\dfrac{211}{1500}$$\\$ $=\dfrac{475+814+211}{1500}$$\\$ $=\dfrac{1500}{1500} =1$$\\$ Therefore, the sum of all these probabilities is $1$.

**3** **In a particular section of Class IX, $40$ students were asked about the months of their birth and the following graph was prepared for the data so obtained:$\\$ Find the probability that a student of the class was born in August.**

Number of students born in the month of August =$ 6$$\\$ Total number of students = $40$$\\$ $P$ (Students born in the month of August) =$\\$ $\dfrac{\text{Number of students born in Au gust }}{\text{Total number of stud ents }}$$\\$ $=\dfrac{6}{40}=\dfrac{3}{20}$

**4** **Three coins are tossed simultaneously $200$ times with the following frequencies of different outcomes:$\\$$\begin{array}{|c|c|}\hline \text{Outcome} & 3 \text{heads}&2 \text{heads} & 1 \text{heads} & \text{No heads}\\ \hline \text{Frequency} & 23 & 72 & 77 & 28 \\ \hline \end{array}$ $\\$ If the three coins are simultaneously tossed again, compute the probability of $2$ heads coming up.**

Number of times $2$ heads come up = $72$$\\$ Total number of times the coins were tossed = $200$$\\$ $P (2$ heads will come up) =$\\$ $\dfrac{\text{Number of times $2$ heads come up}}{\text{Total number of times t he coi ns were tossed }}$$\\$ $=\dfrac{72}{200}=\dfrac{9}{25}$

**5** **An organization selected $2400$ families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:$\\$ Suppose a family is chosen, find the probability that the family chosen is$\\$ (i)earning Rs. $10000-13000$ per month and owning exactly $2$ vehicles.$\\$ (ii)earning Rs. $16000 $or more per month and owning exactly $1$ vehicle.$\\$ (iii)earning less than Rs. $7000$ per month and does not own any vehicle.$\\$ (iv)earning Rs. $13000-16000$ per month and owning more than $2$ vehicles.$\\$ (v)owning not more than $1$ vehicle.**

Number of total families surveyed = $10 + 160 + 25 + 0 + 0 + 305 + 27 + 2 + 1 + 535 + 29 + 1 + 2 + 469 + 59 + 25 + 1 + 579 + 82 + 88 = 2400$$\\$ (i) Number of families earning Rs. $10000-13000$ per month and owning exactly $2$ vehicles = $29$ Hence, required probability,$P=\dfrac{29}{2400}$$\\$ (ii) Number of families earning Rs. $16000$ or more per month and owning exactly $1$ vehicle =$579$$\\$ Hence, required probability,$ P=\dfrac{579}{2400}$$\\$ (iii) Number of families earning less than Rs. $7000$ per month and does not own any vehicle = $10$ Hence, required probability,$P=\dfrac{10}{2400}=\dfrac{1}{240}$$\\$ (iv) Number of families earning Rs. $13000-16000$ per month and owning more than $2$ vehicles =$25$ Hence,required probability,$P=\dfrac{25}{2400}=\dfrac{1}{96}$$\\$ (v) Number of families owning not more than $1$ vehicle = $10 + 160 + 0 + 305 + 1 + 535 + 2 + 469 + 1 + 579 = 2062$ Hence, required probability,$P=\dfrac{2062}{2400}=\dfrac{1031}{1200}$

**6** **A teacher wanted to analyse the performance of two sections of students in a mathematics test of $100$ marks. Looking at their performances, she found that a few students got under $20$ marks and a few got $70 $marks or above. So she decided to group them into intervals of varying sizes as follows:$ 0-20, 20-30, ..., 60-70, 70-100.$ Then she formed the following table:$\\$ $\begin{array}{|c|c|}\hline \text{Mark} & \text{Number of student}\\ \hline 0-20 & 7 \\ \hline 20 - 30 & 10 \\ \hline 30-40 & 10 \\ \hline 40 -50 & 20 \\ \hline 50- 60 & 20 \\ \hline 60-70 & 15 \\ \hline 70-\text{above} & 8 \\ \hline \text{Total} & 90 \\ \hline \end{array}$ $\\$ (i) Find the probability that a student obtained less than $20\%$ in the mathematics test.$\\$ (ii) Find the probability that a student obtained marks $60$ or above.**

Total number of students = $90$ $\\$ (i) Number of students getting less than $20 \%$ marks in the test = $7$ $\\$ Hence, required probability, $P \dfrac{7}{90}$ $\\$ (ii) Number of students obtaining marks $60$ or above = $15 + 8 = 23$ $\\$ Hence, required probability, $P=\dfrac{23}{90}$

**7** **To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.$\\$ $\begin{array}{|c|c|}\hline \text{Option} & \text{Number of students} \\ \hline \text{like} & 135 \\ \hline \text{dislike} & 65 \\ \hline \end{array}$ $\\$ Find the probability that a student chosen at random$\\$ (i) likes statistics$\\$ (ii) does not like it**

Total number of students $= 135 + 65 = 200$ $\\$ (i) Number of students liking statistics = $135 $ $\\$ P (students liking statistics) = $\dfrac{135}{200}=\dfrac{27}{40}$ $\\$ (ii) Number of students who do not like statistics =$ 65$ $\\$ P (students not liking statistics) =$\dfrac{65}{200}=\dfrac{13}{40}$

**8** **The distance (in km) of $ 40 $ engineers from their residence to their place of work were found as follows. $\\$ $ 5 \ \ 3 \ \ 10 \ \ 20 \ \ 25 \ \ 11 \ \ 13 \ \ 7 \ \ 12 \ \ 31 $ $\\$ $ 19 \ \ 10 \ 12 \ \ 17 \ \ 18 \ \ 11 \ \ 32 \ \ 17 \ \ 16 \ \ 2 $ $\\$ $ 7 \ \ 9 \ \ 7 \ 8 \ \ 3 \ \ 5 \ \ 12 \ \ 15 \ \ 18 \ \ 3 $ $\\$ $ 12 \ 14 \ \ 2 \ \ 9 \ \ 6 \ \ 15 \ \ 15 \ \ 7 \ \ 6 \ \ 12 $ $\\$ What is the empirical probability that an engineer lives: $\\$ (i)less than $7 km$ from her place of work ? $\\$ (ii)more than or equal to $7 km$ from her place of work? $\\$ (iii)within $1 /2 km $ from her place of work ?**

(i) Total number of engineers = $40$ $\\$ Number of engineers living less than $7 km$ from their place of work =$ 9$$\\$ Hence, required probability that an engineer lives less than $7 km$ from her place of work, $P=\dfrac{9}{40}$ $\\$ (ii) Number of engineers living more than or equal to $7 km$ from their place of work = $40 - 9 = 31$ $\\$ Hence, required probability that an engineer lives more than or equal to $7 km $ from her place of work, $P =\dfrac{31}{40}$$\\$ (iii) Number of engineers living within $1/2 km$ from her place of work =$ 0$ $\\$ Hence, required probability that an engineer lives within $1/2 km$ from her place of work,$ P = 0$

**9** **Activity Note the frequency of two - wheeler, three-wheeler and four-wheelers going past during a time - interval in front of school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.**

To be done by individual self.

**10** **Activity$\\$ Ask all the students in your class to write a $ 3$-digit number. Choose any student from the room at random. What is the probability that the number written by him/her is divisible by $3$ ? Remember that a number is divisible by $3$, if the sum of it’s digits is divisible by $ 3$.**

To be done by individual self.

**11** **Eleven bags of wheat flour, each marked $5 kg$, actually contained the following weights of flour (in kg):$\\$ $4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00$$\\$ Find the probability that any of these bags chosen at random contains more than $5 kg$ of flour.**

Number of total bags = $11$ $\\$ Number of bags containing more than $5 kg$ of flour = $7$ $\\$ Hence, required probability, $P=\dfrac{7}{11}$

**12** **The below frequency distribution table represents the concentration of Sulphur dioxide in the air in parts per million of a certain city for $30 $ days. Using this table, find the probability of the concentration of Sulphur dioxide in the interval $0.12-0.16$ on any of these days.$\\$ $ \begin{array}{|c|c|}\hline \text{Concentration of $SO_2$ (in ppm)} & \text{Number of days(frequency)}\\ \hline 0.00-0.04 & 4 \\ \hline 0.04-0.08 & 9 \\ \hline 0.08-0.12 & 9 \\ \hline 0.12-0.16 & 2 \\ \hline 0.16 -0.20 & 4 \\ \hline 0.20- 0.24 & 2 \\ \hline \end{array}$**

Number days for which the concentration of sulphur dioxide was in the interval of $0.12-0.16 = 2$ $\\$ Total number of days = $30$ Hence, required probability, $P=\dfrac{2}{30}=\dfrac{1}{15}$

**13** **The below frequency distribution table represents the blood groups of $30$ students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group $AB.$ $\\$ $\begin{array}{|c|c|}\hline \text{Blood group} & \text{Number of students} \\ \hline A & 9 \\ \hline B & 6 \\ \hline AB & 3 \\ \hline O & 12 \\ \hline \text{Total} & 30 \\ \hline \end{array}$**

Number of students having blood group $AB = 3$ $\\$ Total number of students = $30$ $\\$ Hence, required probability, $P=\dfrac{3}{30}=\dfrac{1}{10}$