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Questions  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Q)

Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(a) The youngest is a girl.
(b) At least one is a girl.
Pre-natal *** determination is a crime. What will you do if you come to know that some of your known is indulging in pre-natal *** determination ?


$(A)\; 1/3 , 1/2$
$(B)\; 1/2, 1/3$
$(C)\; 1/4 , 3/4$
$(D)\; 1/4 , 1/2$

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A)
$S = \{ (B,b) , (B,g), (G,b) , (G,g)\}$
B =elder male child,
b = younger male child,
G= elder female child,
g = younger female child
(a) Let E_1 be the event that the youngest is a girl . $E_1 = \{ (Bg) , (Gg)\}$
$P(E_1)= \frac{2}{4} = \frac{1}{2}$
Let $E_2$ be the event that both are girls
$E_2 = \{(Gg)\} = P (E_2)= \frac{1}{4}$
$\therefore E_2 \cap E_1 = \{Gg\} . P(E_2 \cap E_1) = \frac{1}{4}$
$\begin{align*}\therefore P (\frac{E_2}{E_1} ) &= \frac{P (E_2 \cap E_1)}{P(E_1)} \\ & = \frac{1/4}{1/2} \\ & = 1/2 \end{align*}$
(b) Let $E_3 $ be the event that at least one is a girl = $\{(Bg), (Gb), (Gg)\}$
$P(E_3) = 3/4$
$P(E_3 \cap E_2) = \frac{1}{4}$
$\therefore \frac{P(E \cap E_2)}{P(E_3)} = \frac{1/4}{3/4} = 1/3$
Pre-natal determination is a crime. Hence we can make them aware of this crime and educate them not to indulge in such activities.
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