Question

A letter is known to have come either from tatanagar or from calcutta. On the envelope just two consecutive letters ta are visible. What is the probability that the letter came from calcutta

Answer 1

hope ti helps .…... . ............

Answer 2

Probability is $\frac {4}{11}$

Step-by-step explanation:

In the case of TATANAGAR, $E_1$ be an event of letter arrival

&  

In the case of CALCUTTA, $E_2$ be an event of letter arrival

$E_2$ be the event that has come from CALCUTTA.

Let A signify the occasion that the two back to back letters on the envelope are TA.

So, $E_1 & E_2$ are mutually exclusive and exhaustive events.

$P(E_1) = \frac {1}{2}, P(E_2) = \frac {1}{2}$

$P(\frac {A}{E_1})$= Probability that the consecutive letters TA visible on the envelope belong to TATANAGAR.

= $\frac {2}{8}$ (There are 8 consecutive letters namely, TA, AT, TA, AN, NA, AG, GA, AR out of which 2 cases are favorable.)

Similarly, $P(\frac {A}{E_2})$ =  Probability that the consecutive letters TA visible on the envelope belong to CALCUTTA  

= $\frac {1}{7}$(There are seven consecutive letters CA, AL, LC, CU, UT, TT, TA out of which one case is favorable)

So, $P(\frac {E_2}{A}) = \frac {P(E_2)P(\frac {A}{E_2})}{P(E_1).P(\frac {A}{E_1})+P(E_2).P(\frac {A}{E_2})}$

= $\frac {\frac {1}{2}\times \frac {1}{7}}{\frac {1}{2} \times \frac {1}{4} + \frac {1}{2} \times \frac {1}{7}}$

= $\frac {\frac {1}{7}}{\frac {1}{4}+\frac {1}{7}}$

=  $\frac {\frac {1}{7}}{\frac {7+4}{28}}$

= $\frac {1}{7} \times \frac {28}{11}$

= $\frac {4}{11}$