Probability of drawing two or more black balls

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.

• 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. – Alain Remillard Feb 15 '20 at 15:55
• 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. – JMoravitz Feb 15 '20 at 15:58

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 $$P(X \geq 2) = 1- P(X \leq 1) = 1- (P(X = 1) + P(X = 0)) \approx 0.4075$$.

• I have edited the question, 4 balls are drawn. – user121 Feb 15 '20 at 15:59
• $n=12$? No. $n$ would be the number of draws being made, which you just said in the line before you are assuming is $4$. – JMoravitz Feb 15 '20 at 15:59
• My bad, corrected my mistake. – Mevve Feb 15 '20 at 16:02

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.

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