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I need help with some analysis of the following:
I have a jar with 13 balls, 5 red, 3 white and 5 blue.
What is the probability for picking 8 balls and I get 4 red, 2 white and 2 blue, random pick?
What is the thought analsis for solving this?
Thanks,
Mike
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(I'm assuming here that balls don't get replaced after they're picked. If I'm wrong then the numbers need to be changed a bit.)
First, work out the probability of picking 4 reds, then 2 whites, then 2 blues (in order).
For the first pick, there are 5 red balls out of 13 possible ones, so the probability of picking red is 5/13.
For the second, there are 4 reds out of 12, so the probability of picking a 2nd red is 4/12 = 1/3.
Using the same reasoning, the probabilities for the 3rd and 4th reds are 3/11 and 2/10 (=1/5)respectively.
Now we want the probabilities of picking whites. There are 3 white balls out of a total of 9, so the probability of the first white is 3/9 = 1/3.
Similarly, the probability for the second white is 2/8 = 1/4.
Finally, the same reasoning shows that the probabilities for the blue balls are 5/7 and 4/6 (=2/3).
To get the probability of all this happening, we multiply everything together.
5/13 * 1/3 * 3/11 * 1/5 * 1/3 * 1/4 * 5/7 * 2/3 = 150/540540 = 5/18018.
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That's the probability for RRRRWWBB in that order. There are many other orders the would work, however. The total amount is found by 8! / (4!*2!*2!) = 420.
The probability for each of these orders is the same, and so the final answer is (5*420)/18018 = 50/429, or around 1/9.
Why did the vector cross the road?
It wanted to be normal.
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Thank you for your insight. It is very helpful.
Mike
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Z Is A Standard Normal Random Variable
How Do You Do This One
P(z>-2.08)
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