Question

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 :

• A. $\frac{26}{49}$
• B. $\frac{32}{49}$
• C. $\frac{27}{49}$
• D. $\frac{21}{49}$

Answer 1

Answer: (B)

$\frac{32}{49}$

Answer 2

$E_1$ : Event of drawing a Red ball and placing a green ball in the bag
$E_2$ : Event of drawing a green ball and placing a red ball in the bag
E : Event of drawing a red ball in second draw

$P\left(E\right) = P\left(E_{1}\right) \times P\left(\frac{E}{E_{1}} \right) +P\left(E_{2} \right)\times P\left(\frac{E}{E_{2}}\right)$
$= \frac{5}{7} \times \frac{4}{7} + \frac{2}{7} \times \frac{6}{7} = \frac{32}{ 49}$