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Question
A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from
(i) LONDON (ii) CLIFTON?
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Solution
Let A, E1 and E2 denote the events that the two consecutive letters are visible, the letter has come from LONDON and the letter has come from CLIFTON, respectively.
\[\therefore P\left( E_1 \right) = \frac{1}{2} \]
\[ P\left( E_2 \right) = \frac{1}{2}\]
\[\text{ Now } , \]
\[P\left( A/ E_1 \right) = \frac{2}{5}\]
\[P\left( A/ E_2 \right) = \frac{1}{6}\]
\[\text{ Using Bayes' theorem, we get} \]
\[\left( i \right) \text{ Required probability} = P\left( E_1 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{2} \times \frac{2}{5}}{\frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{1}{6}}\]
\[ = \frac{\frac{2}{5}}{\frac{2}{5} + \frac{1}{6}} = \frac{\frac{2}{5}}{\frac{17}{30}} = \frac{12}{17}\]
\[\left( ii \right) \text{ Required probability } = P\left( E_2 /A \right) = \frac{P\left( E_1 \right)P\left( A/ E_1 \right)}{P\left( E_1 \right)P\left( A/ E_1 \right) + P\left( E_2 \right)P\left( A/ E_2 \right)}\]
\[ = \frac{\frac{1}{2} \times \frac{1}{6}}{\frac{1}{2} \times \frac{2}{5} + \frac{1}{2} \times \frac{1}{6}}\]
\[ = \frac{\frac{1}{6}}{\frac{2}{5} + \frac{1}{6}} = \frac{\frac{1}{6}}{\frac{17}{30}} = \frac{5}{17}\]
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`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
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