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.
Best Answer
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.