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  • A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from TATA NAGAR.

A letter is known to have come either from TATA NAGAR or from CALCUTTA. On the envelope, just two consecutive letter TA are visible. What is the probability that the letter came from TATA NAGAR.

Answers (1)

Let events E1, E2 be the following events-

E1 be the event that letter is from TATA NAGAR and E2 be the event that letter is from CALCUTTA

Let E be the event that on the letter, two consecutive letters TA are visible.
Since, the letter has come either from CALCUTTA or TATA NAGAR

\therefore \mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{1}{2}=\mathrm{P}\left(\mathrm{E}_{2}\right)

We get the following set of possible consecutive letters when two consecutive letters are visible in the case of TATA NAGAR
{TA, AT, TA, AN, NA, AG, GA, AR}
We get the following set of possible consecutive letters in the case of CALCUTTA,

{CA, AL, LC, CU, UT, TT, TA}

Therefore, P(E|E1) is the probability that two consecutive letters are visible when letter came from TATA NAGAR
P(E|E2) is the probability that two consecutive letters are visible when letter came from CALCUTTA

\therefore P\left(E \mid E_{1}\right)=\frac{2}{8}, P\left(E \mid E_{2}\right)=\frac{1}{7}

To find- the probability that the letter came from TATA NAGAR.
Using Bayes’ theorem to find the probability of occurrence of an event A when event B has already occurred.

\begin{aligned} &\underset{\therefore}{\mathbf{P}}(\mathbf{A} \mid \mathbf{B})=\frac{P(A) P(B \mid A)}{P(B)}\\ &\mathrm{P}\left(\mathrm{E} \mid \mathrm{E}_{1}\right) \text { is the probability that the letter came from TATA NAGAR }\\ &\therefore P\left(E \mid E_{1}\right)=\frac{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)}{P\left(E_{1}\right) \times P\left(E \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(E \mid E_{2}\right)}\\ &=\frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{2}{8}+\frac{1}{2} \times \frac{1}{7}}\\ &=\frac{\frac{1}{8}}{\frac{1}{8}+\frac{1}{14}}\\ &=\frac{\frac{1}{8}}{\frac{7+4}{56}}=\frac{1}{8} \times \frac{56}{11}=\frac{7}{11} \end{aligned}

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infoexpert22

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