An urn contains 5 red and 2 green balls.

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 :

  1. $\frac{26}{49}$

  2. $\frac{32}{49}$

  3. $\frac{27}{49}$

  4. $\frac{21}{49}$


Correct Option: , 2

Solution:

$\mathrm{E}_{1}$ : Event of drawing a Red ball and placing a green ball in the bag

$\mathrm{E}_{2}$ : Event of drawing a green ball and placing a red ball in the bag

E : Event of drawing a red ball in second draw

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

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

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