Probability of taking balls from a bag

I have a bag that contains balls:

1. 2 blue
2. 3 red
3. 1 white

ball.

My task is to calculate, what is the probability that after 3 rounds of taking bals, i still have no blue balls.

I am not sure if i translated it correctly, but:

P(1B) = probability that i take blue ball in first round

P(2B) = probability that i take blue ball in second round

and so on.

My task is to find $P(1M^{c}\cap{} 2M^{c}\cap3M^{c})$.

Now i know that $P(A^{1}\cap{} A^{2}\cap{}A^{3})$ = $P(A^{1}) * P(A^{2}| A^{1}) * P(A^{3}| A^{1} \cap{} A^{2})$

However in my case i would have to write a lot of things:

$P(1M^{c}) = 4/6$

$P(2M^{c} | 1M^{c})$ = $\frac{P(1M^{c}\cap{}2M^{c})}{1M^{c}}$

$P(1M^{c}\cap{}2M^{c})$ have 3 combinations = (red,red),(red,white),(white,red)

and if i wanted to calculate

$P(3M^{c} | 1M^{c} \cap{} 2M^{c})$ i would have also (red,red,red,),(red,red,white),(red,white,red),(white,red,red)

So far we have 3 combinations, and lot of calculations, how could i find the probability more intuitive? or with some trick, am i missing something?

Thanks for help.

1 Answer

This probability is given by$$\frac{4*3*2}{6*5*4}=\frac{1}{5}=0.2$$Where the numerators are the number of balls we can select each time and the denominators are the number of balls left in the bag.