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What is the probability of drawing with replacement two or more black balls from a hat with 12 balls: four black, four red, and four blue? We are drawing 4 balls from a hat. I have drawn tree diagram. For example, if the first ball drawn is the red one, then we have possible sequences:

Red, Red, Black, Black (probability is $\frac{1}{3^{4}}$ )

Red, Blue, Black, Black

Red, Black, Black (probability is $\frac{1}{3^3}$ )

Red, Black, Blue, Black

Red, Black, Red, Black

so probability for the case when the first ball is red is $4 \frac{1}{3^{4}} + \frac{1}{3^{3}} $

In similar way, I found the probabilities for cases when the first ball is the blue one or black one.

I found that the result is 0.4074. Can someone just check if this is correct? Thanks in advance.

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  • $\begingroup$ The number of draws isn't specify in your question. Do you draw a ball exactly four times? In that case you could use binomial distribution. $\endgroup$ – Alain Remillard Feb 15 '20 at 15:55
  • $\begingroup$ It is going to be much easier on you if you treat blue and red balls as the same category, "not black". This will reduce the case work considerably. $\endgroup$ – JMoravitz Feb 15 '20 at 15:58
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How many draws do you make? From your example I guess 4 but it is not specified in the question.

This could be interpreted as an binomial distribution $X \sim Bin(n=4, p = 1/3)$ so what you should calculate is $P(X \geq 2) = 1- P(X \leq 1) = 1- (P(X = 1) + P(X = 0)) \approx 0.4075$.

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  • $\begingroup$ I have edited the question, 4 balls are drawn. $\endgroup$ – user121 Feb 15 '20 at 15:59
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    $\begingroup$ $n=12$? No. $n$ would be the number of draws being made, which you just said in the line before you are assuming is $4$. $\endgroup$ – JMoravitz Feb 15 '20 at 15:59
  • $\begingroup$ My bad, corrected my mistake. $\endgroup$ – Mevve Feb 15 '20 at 16:02
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You can ease your casework two ways. First, consider the blue and red balls to be nonblack with probability $\frac 23$. You don't need to distinguish them. Second, all different orders of a given combination have the same probability, so compute the chance of one and multiply by the number of different orders.

You can get two black and two nonblack with chance ${4 \choose 2}\left( \frac 13\right)^2\left( \frac 23\right)^2=\frac {24}{81}$
You can get three black and one nonblack with chance ${4 \choose 3}\left( \frac 13\right)^3\left( \frac 23\right)^1=\frac {8}{81}$
You can get four black and no nonblack with chance ${4 \choose 4}\left( \frac 13\right)^4\left( \frac 23\right)^0=\frac {1}{81}$

For a total of $\frac {33}{81}.\ \ $ $0.4074$ is approximately correct. I would leave it as a fraction unless you are asked for a decimal.

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  • $\begingroup$ @N.F.Taussig: Thanks. Fixed. $\endgroup$ – Ross Millikan Feb 15 '20 at 16:32

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