An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
The urn contains 5 red and 5 black balls.
Let a red ball be drawn in the first attempt.
$\therefore P($ drawing a red ball $)=\frac{5}{10}=\frac{1}{2}$
If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.
$P($ drawing a red ball $)=\frac{7}{12}$
Let a black ball be drawn in the first attempt.
$\therefore \mathrm{P}$ (drawing a black ball in the first attempt) $=\frac{5}{10}=\frac{1}{2}$
If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
$P($ drawing a red ball $)=\frac{5}{12}$
Therefore, probability of drawing second ball as red is $\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2} \times 1=\frac{1}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.