Question 1154150
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With this kind of problem, you must specify whether it's "SELECTING WITH
REPLACEMENT" or "SELECTING WITHOUT REPLACEMENT".  You didn't specify I will have
to do it both ways.

If it's WITHOUT replacing, then we will think of reaching in one time and
drawing out 3, so we will use combinations, for you don't care which of the
three balls is drawn first, second or third when you just reach in and get a
handful of 3 balls.  

a) What is the probability that the selected balls have the same color?

WITHOUT REPLACEMENT:

There is only 1 'successful' way, which is to choose all 3 blue balls.

The number of ways to choose ANY three balls is  

6 balls CHOOSE 3 = 6C3 = 20

So the probability is 1 way out of 20 or 1/20 or 0.05 or 5% of the time.

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a) What is the probability that the selected balls have the same color ?

WITH REPLACEMENT:

Case 1: three whites.
The first drawn is white, we do that 1/6 of the time.
The second drawn is white, we also do that 1/6 of the time.
The third drawn is white, we also do that 1/6 of the time.
So (1/6)³ = 1/216 of the time we'll draw 3 whites.

Case 2: three yellows.
The first drawn is yellow, we do that 2/6 or 1/3 of the time.
The second drawn is yellow, we also do that 2/6 or 1/3 of the time.
The third drawn is yellow, we also do that 2/6 or 1/3 of the time.
So (1/3)³ = 1/27 of the time we'll draw 3 yellows.

Case 3: three blues.
The first drawn is blue, we do that 3/6 or 1/2 of the time.
The second drawn is blue, we do that 3/6 or 1/2 of the time.
The third drawn is blue, we do that 3/6 or 1/2 of the time.
So (1/2)³ = 1/8 of the time we'll draw 3 blues.

So the total probability for all three cases is
{{{(1/6)^3+(1/3)^3+(1/2)^3 = 1/216+1/27+1/8=1/216+8/216+27/216 = 36/216 = 1/6}}}

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b) What is the probability that no white balls have been selected?

WITHOUT REPLACING:
That means we draw 3 balls from the 5 non-white balls. 
That's 5 non-white balls CHOOSE 3 or 5C3 = 10 ways.

10 ways out of 20 is 10/20 with reduces to 1/2 
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WITH REPLACING.

c) What is the probability that all [BOTH] yellow balls are selected ?

We pick the two yellows 1 way.
We pick the 1 non-yellow to go with the two yellows in 4 ways.
So there are 1∙4=4 ways to have a combination of 3 including both yellows.

That's 4 ways out of 20 or 4/20 which reduces to 1/5.
 
Edwin</pre>