Question

A bag contains 4 red balls, 6 blue balls and 8 pink balls.
One ball is drawn at random and replace with 3 pink balls.
A probability that the first ball drawn was either red or blue in colour and the second ball drawn was pink in colour?​

Answer 1

Answer:1st 10/18

2nd idk

Step-by-step explanation:

For you should add the. Of balls -.: 4+6=10 (red + blue balls)

Answer 2

Answer:

The probability for the given is $\frac{11}{36}$.

Given:

A bag contains 4 red balls, 6 blue balls and 8 pink balls.One ball is drawn at random and replace with 3 pink balls.

To find:

Probability that the first ball drawn was either red or blue in colour and the second ball drawn was pink in colour?

Solution:  

   Total no of balls = 6 + 4 + 8 = 18 balls

(i) Probability that the first ball is either red or blue = $\frac {(^{4} \mathrm{C}_{1}+ ^{6} \mathrm{C}_{1})}{^{18} \mathrm{C}_{1}} = \frac {10}{18}$

(ii) To replace with 3 pink balls. Total balls = 17 + 3 = 20 balls.

The ball drawn is pink = $\frac{^{11} \mathrm{C}_{1}}{^{20} \mathrm{C}_{1}} = \frac{11}{20}$

Total probability = $\frac{10}{18} + \frac{11}{20} = \frac{11}{36}.$