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I have a list of trials where I pick balls at random from a bag containing different color balls. My focus right now is on a specific color say RED such that

Trial = {RED, NoRED, RED, NoRED, NoRED, NoRED, NoRED, RED, NoRED, NoRED}
From this data N = 10, p=numberOfRed/n = 3/10 = 0.3, q=1-p=0.7

Now I have a new data point. What is the probability of the ball being RED? Here is how I am trying to solve this problem.

P(X=RED) = N!/(N-X)!*X!*(p^N*q^(N-X))
P(X=RED) = 10!/(10-1)!*1!*(0.3^10*0.7^9)

In this case, for a new data point I am putting X=1. I think that is not correct. Can somebody help in calculating the probability of a new data point being a RED ball ? Thanks

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    $\begingroup$ Math typesetting improves readability and the probability of receiving answers. More info: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Sycorax
    Sep 14, 2017 at 23:36
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    $\begingroup$ Selecting balls from bags suggests the possibility that you're sampling without replacement. Are you? $\endgroup$
    – Glen_b
    Sep 15, 2017 at 2:11

2 Answers 2

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Your equation for P(X) will tell you the probability of getting a number of red balls equal to X if you perform N trials. So if you want to use those equations you should also assume N = 1. But, it's simpler just to see that the estimated probability is p. If you work out P(X=1) for N=1, you'll see that you just get P(X=1) = p.

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It's very important when you're given equations to understand what they mean, and to not just blindly plug numbers in. Otherwise you just wind up wasting a lot of time, and more importantly, you learn nothing.

You performed 10 trials and observed 3 red balls. Thus your best estimate of the "true" probability of sampling a red ball is 3/10=0.3. That's.... it. That's the probability.

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