# Probability

## Class 9 NCERT Maths

### NCERT

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 }}$$\\ =\dfrac{24}{30}=\dfrac{4}{5} 2 1500 families with 2 children were selected randomly, and the following data were recorded: Compute the probability of a family, chosen at random, having\\ \begin{array}{|c|c|}\hline \text{Number of girls in a family} & 2 & 1 & 0 \\ \hline \text{Number of families} & 475 & 814 & 211 \\ \hline \end{array} \\ (i)2 girls\\ (ii)1 girl\\ (iii)No girl\\ Also check whether the sum of these probabilities is 1. ##### Solution : Total number of families = 475 + 814 + 211 = 1500$$\\$ (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.

##### Solution :

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.

##### Solution :

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$(in ppm)} & \text{Number of days(frequency)}\\ \hline 0.00-0.04 & 4 \\ \hline 0.04-0.08 & 9 \\ \hline 0.08-0.12 & 9 \\ \hline 0.12-0.16 & 2 \\ \hline 0.16 -0.20 & 4 \\ \hline 0.20- 0.24 & 2 \\ \hline \end{array}$

##### Solution :

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 $AB.$ $\\$ $\begin{array}{|c|c|}\hline \text{Blood group} & \text{Number of students} \\ \hline A & 9 \\ \hline B & 6 \\ \hline AB & 3 \\ \hline O & 12 \\ \hline \text{Total} & 30 \\ \hline \end{array}$

##### Solution :

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