In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

# Ex.15.1 Q2 Probability Solution - NCERT Maths Class 9

Go back to  'Ex.15.1'

## Question

$$1500$$ families with $$2$$ children were selected randomly, and the following data were recorded:

 Number of girls in a family $$2$$ $$1$$ $$0$$ Number of families $$475$$ $$814$$ $$211$$

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 $$1.$$

Video Solution
Probability
Ex exercise-15-1 | Question 2

## Text Solution

What is known?

Number of families having $$2$$ girl child,$$1$$ girl child, no girl child and total number of families.

What is the unknown?

Probability of selecting family with $$2$$ or $$1$$ or $$0$$ girls.

Reasoning:

The empirical probability $$P(E)$$ of an event $$E$$ happening, is given by:

\begin{align}{P}({E})=\frac{ \begin{bmatrix} \text { Number of trials in which }\\ \text{the event happened } \end{bmatrix} }{ \begin{bmatrix} \text { The total number} \\ \text{of trials } \end{bmatrix} }\end{align}

Use probability to derive the solution where

Probability of an selecting family

\begin{align} & { = \frac{{\text{Family having 2 or 1 }} {\text{or 0 girl child}} } {{{\text{Total number of families}}}}} \end{align}

Steps:

No of families having $$2$$ girl child $$= 475$$

No of families having $$1$$ girl child $$= 814$$

No of families having $$0$$ girl child $$= 211$$

Therefore, Total number of families $$= 1500$$

Probability $$P1$$ [family having $$2$$ girl child]

\begin{align} & { = \frac{{\text{Family having }} {\text{ 2 girl child}} } {{{\text{Total number of families}}}}} \end{align}

\begin{align}\ \therefore Pl=\frac{475}{1500}=\frac{19}{60}\end{align}

Probability $$P2$$ [family having $$1$$ girl child]

\begin{align} & { = \frac{{\text{Family having }}{\text{ 1 girl child}} } {{{\text{Total number of families}}}}} \end{align}

\begin{align} \therefore P2=\frac{814}{1500}=\frac{407}{750}\end{align}

Probability $$P3$$ [family having $$0$$ girl child]

\begin{align} & { = \frac{{\text{Family having }} {\text{ 0 girl child}} } {{{\text{Total number of families}}}}} \end{align}

\begin{align} \therefore P3=\frac{211}{1500}\end{align}

Sum of all the three probabilities

\begin{align} &={P1}+{P2} +{P3} \\ &=\frac{475}{1500}+\frac{814}{1500}+\frac{211}{1500} \\ &=\frac{1500}{1500} \\ &=1 \end{align}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school