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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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प्रश्न

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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उत्तर

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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अध्याय 30: Probability - Exercise 31.7 [पृष्ठ ९६]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 30 Probability
Exercise 31.7 | Q 8 | पृष्ठ ९६

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