A letter is known to have arrived by post either from KANPUR or from ANANTPUR.

Question

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

Accepted Answer

Let

$A$ = event that the letter came from ANANTPUR

$K$ = event that the letter came from KANPUR

$B$ = event that on the envelope, the two visible consecutive letters are AN

We have to find:

$P(A|B)$

Using Bayes' theorem,

$P(A|B)=\frac{P(A)\, P(B|A)}{P(A)\, P(B|A)+P(K)\, P(B|K)}$

Since the letter is known to have come either from ANANTPUR or KANPUR, and no other information is given, we take

$P(A)=P(K)=\frac{1}{2}$

So,

$P(A|B)=\frac{P(B|A)}{P(B|A)+P(B|K)}$

Now we find these probabilities.

1. Probability of seeing AN in ANANTPUR
Word: ANANTPUR

Its consecutive pairs are:

$AN,NA,AN,NT,TP,PU,UR$

There are $7$ consecutive pairs in all, and AN occurs $2$ times.

Therefore,

$P(B|A)=\frac{2}{7}$

2. Probability of seeing AN in KANPUR
Word: KANPUR

Its consecutive pairs are:

$KA,AN,NP,PU,UR$

There are $5$ consecutive pairs in all, and AN occurs $1$ time.

Therefore,

$P(B|K)=\frac{1}{5}$

3. Apply Bayes' theorem
$P(A|B)=\frac{\frac{2}{7}}{\frac{2}{7}+\frac{1}{5}}$

Take LCM:

$\frac{2}{7}+\frac{1}{5}=\frac{10+7}{35}=\frac{17}{35}$

So,

$P(A|B)=\frac{\frac{2}{7}}{\frac{17}{35}}=\frac{2}{7}\cdot \frac{35}{17}=\frac{10}{17}$

Therefore, the required probability is

$\frac{10}{17}$

So, the correct option is B.

Solutions · 1