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Ex.15.1 Q2 Probability Solution - NCERT Maths Class 9

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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

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

  
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