Question

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 balls is drawn at random. What is the probability that the second ball is red?

Answer 1

The urn contains 5 red and 5 black balls.

i.e., $n\left(R\right)=5,n\left(B\right)=5andn\left(S\right)=10$

Let a red ball be drawn in the first attempt

$\therefore$ P(drawing a red ball) = $\frac{n\left(R\right)}{n\left(S\right)}=\frac{5}{10}=\frac{1}{2}$

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls i.e., $n\left(R\right)=7,n\left(B\right)=5andn\left(S\right)=12$

P (drawing a red ball)= $\frac{n\left(R\right)}{n\left(S\right)}=\frac{7}{12}$

Let a black ball be drawn in the first attempt

Then, $n\left(R\right)=5,n\left(B\right)=5andn\left(S\right)=10$

$\therefore$ P(drawing a black ball in the first attempt)= $\frac{n\left(B\right)}{n\left(S\right)}=\frac{5}{10}=\frac{1}{2}$

If two blakc balls are added to the urn, then the urn contains 5 red and 7 black balls.

i.e., n(R)=5,n(B)=7 and n(S)=12 $\therefore$ P(drawing a red ball)= $\frac{n\left(R\right)}{n\left(S\right)}=\frac{5}{12}$

Therefore, probability of drawing second balls as red is

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