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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 urn. Now, a second ball is drawn at random from it. The probability that the second ball is red is

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\frac{21}{49}

\frac{27}{49}

\frac{26}{49}

\frac{32}{49}

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detailed solution

Correct option is D

$E_{1} =$ Red ball drawn $P (E_{1}) = \frac{5}{7}$

$E_{2} =$ Red ball drawn $P (E_{2}) = \frac{2}{7}$

A=finally red ball drawn

$\begin{array}{rcl} P (\frac{A}{E_{1}}) & = & \frac{4}{7} P (\frac{A}{E_{2}}) = \frac{6}{7} \end{array}$

$\begin{array}{rcl} P (A) & = & \frac{5}{7} \times \frac{4}{7} + \frac{2}{7} \times \frac{6}{7} = \frac{32}{49} \end{array}$

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