A letter is known to have come from either TATAPUR or from CHAKRATA. On the envelope, only two letters 'TA' are visible consecutively. The probability that the letter has come from CHAKRATA is:

A letter came from either TATAPUR or CHAKRATA. Only two consecutive letters "TA" are visible on the envelope. This requires conditional probability using Bayes' Theorem.

---

TATAPUR has 7 letters: T-A-T-A-P-U-R

Consecutive "TA" appears at:

• Position 1-2: TA
• Position 3-4: TA

TATAPUR contains 2 consecutive "TA" pairs.

---

CHAKRATA has 8 letters: C-H-A-K-R-A-T-A

Consecutive "TA" appears at:

• Position 7-8: TA

CHAKRATA contains 1 consecutive "TA" pair.

---

For a word with $n$ letters, there are $(n-1)$ possible consecutive letter pairs.

TATAPUR: 7 letters → 6 possible pairs

CHAKRATA: 8 letters → 7 possible pairs

---

$P(\text{seeing TA} \mid \text{from TATAPUR})$

$= \dfrac{2}{6}$

$= \dfrac{1}{3}$

$P(\text{seeing TA} \mid \text{from CHAKRATA})$

$= \dfrac{1}{7}$

---

Assuming no prior information:

$P(\text{from TATAPUR}) = \dfrac{1}{2}$

$P(\text{from CHAKRATA}) = \dfrac{1}{2}$

---

$P(\text{TA visible})$

$= P(\text{TA} \mid \text{TATAPUR}) \times P(\text{TATAPUR}) + P(\text{TA} \mid \text{CHAKRATA}) \times P(\text{CHAKRATA})$

$= \dfrac{1}{3} \times \dfrac{1}{2} + \dfrac{1}{7} \times \dfrac{1}{2}$

$= \dfrac{1}{6} + \dfrac{1}{14}$

$= \dfrac{7}{42} + \dfrac{3}{42}$

$= \dfrac{10}{42}$

$= \dfrac{5}{21}$

---

$P(\text{CHAKRATA} \mid \text{TA visible})$

$= \dfrac{P(\text{TA} \mid \text{CHAKRATA}) \times P(\text{CHAKRATA})}{P(\text{TA visible})}$

$= \dfrac{\dfrac{1}{7} \times \dfrac{1}{2}}{\dfrac{5}{21}}$

$= \dfrac{\dfrac{1}{14}}{\dfrac{5}{21}}$

$= \dfrac{1}{14} \times \dfrac{21}{5}$

$= \dfrac{21}{70}$

$= \dfrac{3}{10}$

Therefore, the probability that the letter came from CHAKRATA is $\dfrac{3}{10}$.