1500 families with 2 children were selected randomly, and the following data was recorded:
$\begin{matrix} Number of girls in a family & 2 & 1 & 0 Number of families & 475 & 814 & 211 \end{matrix}$

Compute the probability of a family, chosen at random, having
(i) 2 girls.
(ii) 1 girl.
(iii) no girl.

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Solution

Total number of families = 475 +814 + 211= 1500
(i) Number of families having 2 girls = 475
$P_{1} (a randomly chosen family has 2 girls) = \frac{Number families having 2 girls}{Total number of families}$
$P_{1} (a randomly chosen family has 2 girls) = \frac{475}{1500} = \frac{19}{60}$

(ii) Number of families having 1 girl = 814
$P_{2} (a randomly chosen family has 1 girl) = \frac{Number families having 1 girl}{Total number of families}$
$P_{2} (a randomly chosen family has 1 girl) = \frac{814}{1500} = \frac{407}{750}$

(iii) Number of families having no girl= 211
$P_{3} (a randomly chosen family has no girl) = \frac{Number families having no girl}{Total number of families}$
$P_{3} (a randomly chosen family has no girl) = \frac{211}{1500}$

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Similar questions

\begin{matrix} Number of girls in a family & 2 & 1 & 0 Number of families & 475 & 814 & 211 \end{matrix}

\begin{matrix} N u m b e r o f g i r l s i n a f a m i l y & 2 & 1 & 0 N u m b e r o f f a m i l i e s & 475 & 814 & 211 \end{matrix}

\begin{matrix} N u m b e r o f g i r l s i n a f a m i l y & 2 & 1 & 0 N u m b e r o f f a m i l i e s & 475 & 814 & 211 \end{matrix}

1500

2

No. of girls in a family	$2$	$1$	$0$
No. of families	$475$	$814$	$211$

2

1

1

\begin{matrix} N u m b e r o f b o y s i n a f a m i l y & 0 & 1 & 2 N u m b e r o f f a m i l i e s & 475 & 814 & 211 \end{matrix}

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