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

The correct option is B $\frac{32}{49}$

$E_1$ : 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

$P\left(E\right)=P\left(E_1\right)\times P\left(\frac{E}{E_1}\right)+P\left(E_2\right)\times P\left(\frac{E}{E_2}\right)$

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