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

## 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 \(1.\)

## 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

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