**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$\\$ (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:$\\$ 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,$P=\dfrac{2062}{2400}=\dfrac{1031}{1200}$