Question

There are 3 red and 8 green balls in a bag. If you randomly choose balls one at a time, with replacement, what is the probability of choosing 2 red balls and then 1 green ball?without replacement4/16524/6051/1072/1331

Answer 1

So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a $\frac{3}{11}$ chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is $\frac{8}{11}$, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:

$\frac{3}{11}\frac{3}{11}\frac{8}{11}=\frac{3*3*8}{11^3}=\frac{72}{1331}$

Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is$\frac{2}{11}$, and the probability of picking the second red ball is $\frac{1}{10}$ and the probability of picking the green ball is$\frac{1}{9}$. We want to multiply thm again, as per the multiplication rule like the last problem.

$\frac{2}{11}*\frac{1}{10}*\frac{1}{9}=\frac{1}{11*5*9}=\frac{1}{495}$