A bag contains 4 Red and 6 Black balls. A ball is drawn at random from the bag, its colour is observed and…
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 3 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
$\frac{41}{65}$
$\frac{24}{65}$
$\frac{26}{65}$
$\frac{28}{65}$
Solution
i. Probability that the first ball is black and second is red.
Total number of black balls $=6$
Total number of red balls $=4$
Probability of getting black ball in first draw $=\frac{6}{10}$.
Now, number of black balls $=9$ and Total number of balls = 13
$\therefore \quad$ Probability of getting red ball in second draw $=\frac{4}{13}$.
ii. Probability that both the balls are red. Probability of getting red ball in first draw $=\frac{4}{10}$.
Now for second draw, -
Number of red balls $=7$ and
Total number of balls $=13$.
$\therefore \quad$ Probability of getting red ball in second draw $=\frac{7}{13}$
$\therefore \quad$ Total probability of drawing red ball
$\begin{aligned}
& =\left(\frac{6}{10} \times \frac{4}{13}\right)+\left(\frac{4}{10} \times \frac{7}{13}\right) \\
& =\frac{24+28}{130} \\
& =\frac{26}{65}
\end{aligned}$