Probability that none of the four balls removed from the bag is blue with and without replacement Can you please help me with these questions?
Event 1:
3 red balls, 2 yellow balls, 1 blue ball in a bag
1. take out 1 ball, take out another 1 ball
2. again, take out 1 ball, take out another 1 ball
What's the probability that none of the removed balls is blue?
Here is how I see the question:
The possibility of the first take is: $5/6$ ($5$ non-blue balls of $6$ balls in the bag)
the possibility of the second take is: $3/4$  ($3$ non-blue balls of $4$ balls left in the bag)
so the whole possibility is: $5/6 \cdot 0.5 + 3/4 \cdot 0.5 \approx 0.79$
Event 2:
Still, 3 red balls, 2 yellow balls, 1 blue ball in a bag
1. take out 1 ball, put the ball back
2. take out 1 ball, put the ball back
3. take out 1 ball, put the ball back
4. take out 1 ball, put the ball back       
What's the probability that none of the removed balls is blue?
Here is how I see the question:
Each time taking out a non-blue ball, the chance is $5/6$, so the possibility of the whole event is:
 $5/6 \cdot 0.25 \cdot 4 \approx 0.83$
Am I getting them right?
Thank you.
 A: Event 1:
You are removing 4 balls from 6 without replacement.
there are $\binom 6 4 =15$ combinations that do this without any restriction
If all 4 balls must be chosen from the 5 non-blue balls there are only $\binom 5 4=5 $ combinations
Each combination is equally probable.
Event 2: 
With replacement there are always 6 balls to choose from so the number of permutations for 4 choices is $6^4 = 1296$
but only $5^4=625$ of these have all 4 balls non-blue 
Each permutation is equally probable
A: Nope.   There was no reason to divide by half, and certainly not to add.  Recall the Product Rules on Probabilities for Intersections of Events; in this case the events are dependent.
The probability for intersection of events in (1) is : $\dfrac 56\cdot\dfrac 45$.    Notice, if you extract 1 non-blue ball from the bag, there are 1 blue and 4 non-blue balls left.
(  Also, there are $\dfrac{5\cdot 4}{2\cdot 1}$ ways you draw $2$ from $5$ non-blue balls, and $\dfrac{6\cdot 5}{2\cdot 1}$ ways to draw $2$ from all $6$ balls.   Thusly dividing and cancelling gives the same answer.  )

Evaluating the measure for intersection of events in (2) proceeds in the same manner.   Try your hand at it now.
A: 
What is the probability that if four balls are removed from a bag containing three red balls, two yellow balls, and one blue ball without replacement that none of the removed balls is blue?

Since five of the six balls in the bag are not blue, the probability that the first ball removed from the bag is not blue is $5/6$.  If the first ball removed from the bag is not blue, we are left with five balls in the bag, of which four are not blue.  Hence, the probability that the second ball removed from the bag is also not blue is $4/5$.  By similar reasoning, if the first two balls removed from the bag are not blue, then the probability the third ball removed from the bag is not blue is $3/4$.  If the first three balls removed from the bag are not blue, then the probability the fourth ball removed from the bag is not blue is $2/3$.  Hence, the probability that all four of the balls removed from the bag are not blue is
$$p = \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$$

What is the probability that if four balls are removed from a bag containing three red balls, two yellow balls, and one blue ball with replacement that none of the removed balls is blue?

As before, the probability that the first ball removed from the bag is not blue is $5/6$.  Once that ball is replaced, we are left with six balls in the bag, of which five are not blue.  Hence, the probability that the second ball is also not blue is $5/6$.  The same reasoning applies to the third and fourth draws, so the probability that none of the balls removed from the bag are blue is 
$$\left(\frac{5}{6}\right)^4 = \frac{625}{1296}$$
