First slide

Theorems of probability

Question

$T w o c o i n s a r e a v a i l a b l e, o n e f a i r a n d t h e o t h e r t w o h e a d e d . C h o o s e a c o i n a n d t o s s i f o n c e a s s u m e t h a t t h e u n b i a s e d c o i n i s c h o s e n w i t h p r o b a b i l i t y \frac{3}{4} . G i v e n t h a t t h e o u t c o m e i s h e a d, t h e p r o b a b i l i t y t h a t t h e t w o - h e a d e d c o i n w a s c h o s e n i s$

Moderate

Solution

$P (F) = \frac{3}{4}, P (T) = 1 - \frac{3}{4} = \frac{1}{4}$
$F \to Fair coin, T \to Two headed, H \to Head occurs$
$∴ P (T / H) = \frac{P (H / T) \cdot P (T)}{P (H / T) \cdot P (T) + P (H / F) \cdot P (F)}$
$[Using Baye's theorem]$
$= \frac{1 \cdot \frac{1}{4}}{1 \cdot \frac{1}{4} + \frac{1}{2} \cdot \frac{3}{4}} = \frac{2}{5}$

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