Probability of picking r red balls out of 100

Your example works well with only 3 balls. Let's pretend we have 7 total balls in an urn, 3 red, 4 blue. Each draw is done such that the ball selected is not returned to the urn.

On our first draw we have a 3/7 chance of selecting red, and a 4/7 chance of selecting blue.

Question: What is the probability of selecting a red ball on the 3rd draw?

Answer:

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_____________________

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(3/7) (4/7)

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_____________ ______________

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(2/6) (4/6) (3/6) (3/6)

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_________ _______ _________ _______

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(1/5) (4/5) (2/5) (3/5) (2/5) (3/5) (3/5) (2/5

The probability of selecting a red ball on the 3rd draw is:

((3/7)*(2/6)*(1/5)) + ((3/7)*(4/6)*(2/5)) + ((4/7)*(3/6)*(2/5)) + ((4/7)*(3/6)*(3/5))

This represents choosing RRR + RBR + BRR + BBR.

Wow, that was an utter waste of time since the formatting didn't hold.

precisely wrote:

Wow, that was an utter waste of time since the formatting didn't hold.

Your formatting wasn't an utter waste (I can see it by pushing the QUOTE button), your framing of the problem is an utter waste. You're stuck in a deadend. Take a break and come back when your head is clear.

This is not a problem where you are going down nodes of a binomial tree. It's straightforward probability. The calculations are no different than poker calculations.

Stats Help wrote:

I need some help on stats. So there are 100 balls, r of which are red. What is the probability that the 50th and last balls drawn will be red, without replacement. Thanks.

In truth --pretty slim I would say.

In D=5 the red ballon takes off and the threadmill of LetsRun goes on.

malmo wrote:

Don't forget to bracket the 2nd half of that formula correctly, otherwise it takes on a new meaning.

(r/d)*((r-1)/(d-1))

That did slip - but the calculation can be written with or without the brackets. But I know what you're saying - it REPRESENTS the two probabilities of the relative events that are in question.