Answer to Question #335860 in Statistics and Probability for Mia

Question

There is a bag filled with 3 blue, 2 red and 4 green marbles. A marble is taken at random from the bag, the colour is noted and then it is not replaced. Another marble is taken at random. What is the probability of getting 2 different colours?

Answer 1

We consider two possibilities:

1. The order of marbles does not matter. I.e., it does not matter, which marble was taken first and which marble was taken second. There are: $C_9^2=\frac{9\cdot8}2=36$ different ways to choose $2$ marbles from $9$. Denote by the capital letter a color of the marble. The following combinations correspond to two different colors: $BR, RB,BG,GB, RG,GR.$We get: $\frac{6+6}2=6$ ways to choose a blue and a red marble, $\frac{12+12}{2}=12$ ways to choose a blue and a green marble and $\frac{8+8}{2}=8$ ways to choose a red and a green marble. A division by $2$ is present, since the order of marbles does not matter. The probability to get two marbles with different colors is: $\frac{6+12+8}{36}=\frac{26}{36}=\frac{13}{18}$
2. The order of marbles matters. There are $9\cdot8=72$ ways to choose $2$ marbles from $9$. The following combinations correspond to two different colors: $BR, RB,BG,GB, RG,GR$. Find the respective probabilities for these combinations: $P(BR)=P(RB)=\frac{6}{72}$, $P(BG)=P(GB)=\frac{12}{72}$, $P(RG)=P(GR)=\frac{8}{72}$. The probability is: $p=\frac{12+24+16}{72}=\frac{52}{72}=\frac{13}{18}$

The answer is: $\frac{13}{18}.$