Probability — Class 12 Maths Solution

Let ${E_1}$ be the event that letter is from TATA NAGAR and ${E_2}$ be the event that letter is from CALCUTTA.

Also, let ${E_3}$ be the event that on the letter, two consecutive letters TA are visible.

$\therefore$ $P\left( {{E_1}} \right) = \frac{1}{2}$

and $P\left( {{E_2}} \right) = \frac{1}{2}$

and $P\left( {{E_3}/{E_1}} \right) = \frac{2}{8}$

and $P\left( {{E_3}/{E_2}} \right) = \frac{1}{7}$

[since, if letter is from TATA NAGAR,

we see that the events of two consecutive letters visible are

$\{ {\rm{TA}},{\rm{AT}},{\rm{TA}},{\rm{AN}},{\rm{NA}},{\rm{AG}},{\rm{GA}},{\rm{AR}}\} .$

So, $P\left( {{E_3}/{E_1}} \right) = \frac{2}{8}$

and if letter is from CALCUTTA,
we see that the events of two consecutive letters to visible are

$\{ {\rm{CA}},{\rm{AL}},{\rm{LC}},{\rm{CU}},{\rm{UT}},{\rm{TT}},{\rm{TA}}\} .$

So, $P\left( {{E_3}/{E_2}} \right) = \frac{1}{7}$]
$\therefore$ $\quad P\left( {{E_1}/{E_3}} \right) = \frac{{P\left( {{E_1}} \right) \cdot P\left( {{E_3}/{E_1}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {{E_3}/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {{E_3}/{E_2}} \right)}}$

$= \frac{{\frac{1}{2} \cdot \frac{2}{8}}}{{\frac{1}{2} \cdot \frac{2}{8} + \frac{1}{2} \cdot \frac{1}{7}}} = \frac{{\frac{1}{8}}}{{\frac{1}{8} + \frac{1}{{14}}}} = \frac{{1/8}}{{\frac{{22}}{{8 \times 14}}}} = \frac{{\frac{1}{8}}}{{\frac{{11}}{{56}}}} = \frac{7}{{11}}$