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?
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?

Solution:

Given: Urn contains 5 red and 5 black balls.

Let in first attempt the ball drawn is of red colour.

\Rightarrow P (probability of drawing a red ball) =5 / 10=1 / 2

After the two balls of the same colour (red) have been added to the urn, the urn will contain seven red balls and five black balls.

\Rightarrow P (probability of drawing a red ball) =7 / 12

Now let in first attempt the ball drawn is of black colour.

\Rightarrow P (probability of drawing a black ball) =5 / 10=1 / 2

Now the two balls of same colour (black) are added to the urn then the urn contains 5 red and 7 black balls.

\Rightarrow \mathrm{P} (probability of drawing a red ball) =5 / 12

As a result, the likelihood of drawing the second ball as being of the colour red is as follows:

=\left(\frac{1}{2} \times \frac{7}{12}\right)+\left(\frac{1}{2} \times \frac{5}{12}\right)=\frac{1}{2}\left(\frac{7}{12}+\frac{5}{12}\right)=\frac{1}{2} \times 1=\frac{1}{2}