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An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the um. Now a second ball is drawn at random from it, The probability that the second ball is red, is

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27/49

32/49

21/49

26/49

answer is B.

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Detailed Solution

Let $E_{1}$ be the event of drawing a red ball and adding a green ball, $E_{2}$ be the event of drawing a green ball and adding a red ball and $E$ be the event of drawing a red ball in second draw.

$∴ P (E) = P (E_{1}) \times P (\frac{E}{E_{1}}) + P (E_{2}) \times P (\frac{E}{E_{2}})$

$= \frac{5}{7} \times \frac{4}{7} + \frac{2}{7} \times \frac{6}{7} = \frac{20 + 12}{49} = \frac{32}{49}$

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