Question

A letter is known to have come form TATANAGAR or CALCUTTA. On the envelope just two consecutive letters TA are visible. The probability that the letter has come from CALCUTTA is

• A. $\frac{4}{11}$
• B. $\frac{7}{11}$
• C. $\frac{1}{22}$
• D. $\frac{21}{22}$

Answer 1

The correct option is A $\frac{4}{11}$

Let's calculate the total probability of getting $TA$
$P\left(TA\right)=P\left(TA/TATANAGAR\right)\times P\left(TATANAGAR\right)+P\left(TA/CALCUTTA\right)\times P\left(CALCUTTA\right)TATANAGARareTA,AT,TA,AN,NA,AG,GA,AR$

Now $P\left(TA/TATANAGAR\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)/\left(\begin{array}{c}8\\ 1\end{array}\right)=2/8$

and $P\left(TA/CALCUTTA\right)=1/\left(\begin{array}{c}7\\ 1\end{array}\right)=1/7$

since both CALCUTTA and TATANAGAR are equally likely

$\Rightarrow P\left(TA\right)=1/2\times \{2/8+1/7\}=11/56$

Hence by Bayes' theorem :
$P\left(CALCUTTA/TA\right)=\frac{P\left(TA/CALCUTTA\right)P\left(CALCUTTA\right)}{P\left(TA\right)}=\frac{1}{7}\times \frac{1}{2}\times \frac{56}{11}=\frac{4}{11}$