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A bag contains 6 red balls, 9 green balls, and 4 yellow balls. Five balls are selected from the bag.

What is the probability of selecting 4 yellow balls?

I do know that the total # of outcome is 11,628

so would the # of yellow balls selected for the bag be 126

What really matters is that there are 4 balls that are yellow and 6+9 = 15 blocks that are not.

This is a total of 4+15 = 19 balls.

If 4 yellow balls are selected and 1 one ball that is not,

there is only 1 way to select 4 yellow balls and 15 ways to select one of another color.

That makes for a total of 1*15 = 15 ways to get 4 yellow balls.

To select 5 balls our of a total of 19, the number of possibilities is

19!/(5!(19-5)!) = 19*18*17*16*15/(5*4*3*2*1).

It can be seen that 5*3 = 15, and that reduces the problem to 19*18*17*16/(4*2).

Since 4*2 = 8, and 16/8 = 2, that reduces to 19*18*17*2.

Now I know that 19*18 = 342 and 17*2 = 34.

We have 342*34 = 10,260 + 1,368 = 11,628 total outcomes, and I see that was what was given.

The odds of getting 4 yellows in 5 draws is then 15/11,628, which is roughly 0.00129.

In other words, a little more than one tenth of one percent.

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