A family has 5 children, 2 children are selected at random and are found to be girls. What is the probability that the family has 3 girls and 2 boys?

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Solution

$O u t o f 5 c h i l d r e n w e k n o w t h a t 2 a r e g i r l s . S o t h e p r o b a b i l i t y o f f a m i l y h a v i n g 3 g i r l s a n d 2 b o y s w i l l b e s a m e a s t h a t a m o n g t h e r e m a i n i n g 3 c h i l d r e n, 1 i s a g i r l a n d 2 a r e b o y s L e t u s d e f i n e f o l l w i n g e v e n t s B : A c h i l d i s a b o y G : A c h i l d i s a g i r l E : F a m i l y h a s 3 g i r l s a n d 2 b o y s S i n c e t h e r e r e t w o p o s s i b i l i t i e s e i t h e r g i r l o r a b o y, o u t o f w h i c h o n e i s f a v o u r b l e T h e r e f o r e P (B) = \frac{1}{2} = P (G) N o t e : B a n d G a r e i d e p e n d e n t e v e n t s P (E) = P (a m o n g t h e r e m a i n i n g 3 c h i l d r e n, 1 i s a g i r l a n d 2 a r e b o y s) = P (F \cap B \cap B) + P (B \cap F \cap B) + P (B \cap B \cap F) = P (F) P (B) P (B) + P (B) P (F) P (B) + P (B) P (B) P (F) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = 3 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{3}{8}$

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