Question

a bag contains 4 red and 6 black balls .a ball is drawn at random from the bag its colour is observed and this ball along with two additional balls of the same colour are returned to the bag if now a ball is drawn at random from the bag then the probability that this drawn ball is red is

Answer 1

(4/10)(6/12)+(6/10)(4/12)=2/5
hope u will understand

Answer 2

The probability of red ball drawn is $\frac{2}{5}$

Solution:

Probability of red balls: $\frac{4}{10}$

Probability of black balls: $\frac{6}{10}$

Let $X_{i}$ be the event in which a red ball is drawn for the $i^{th}$ term and $Y_{i}$be the event in which a black ball is drawn for the $i^{t h}$ term. Now in the bag there are all total 10 balls i.e. 4 red balls and 6 black balls.  

$\mathrm{P}\left(X_{i}\right)=\frac{4}{10} \text { and } \mathrm{P}\left(Y_{i}\right)=\frac{6}{10}$

As given in the question,two balls drawn extra in each colors, $P\left(\frac{X_{2}}{X_{1}}\right)=\frac{6}{12} \text { and } P\left(\frac{X_{2}}{Y_{1}}\right)=\frac{4}{12}$

Hence the required probability for the chances to get a red ball is  $\bold{\mathrm{P}\left(X_{i}\right) P\left(\frac{X_{2}}{X_{1}}\right)+\mathrm{P}\left(Y_{i}\right) P\left(\frac{X_{2}}{Y_{1}}\right)=\frac{4}{10} \cdot \frac{6}{12}+\frac{6}{10} \cdot \frac{4}{12}=\frac{2}{5}}$