Question #335860

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?


1
Expert's answer
2022-05-03T16:51:10-0400

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: C92=982=36C_9^2=\frac{9\cdot8}2=36 different ways to choose 22 marbles from 99. Denote by the capital letter a color of the marble. The following combinations correspond to two different colors: BR,RB,BG,GB,RG,GR.BR, RB,BG,GB, RG,GR.We get: 6+62=6\frac{6+6}2=6 ways to choose a blue and a red marble, 12+122=12\frac{12+12}{2}=12 ways to choose a blue and a green marble and 8+82=8\frac{8+8}{2}=8 ways to choose a red and a green marble. A division by 22 is present, since the order of marbles does not matter. The probability to get two marbles with different colors is: 6+12+836=2636=1318\frac{6+12+8}{36}=\frac{26}{36}=\frac{13}{18}
  2. The order of marbles matters. There are 98=729\cdot8=72 ways to choose 22 marbles from 99. The following combinations correspond to two different colors: BR,RB,BG,GB,RG,GRBR, RB,BG,GB, RG,GR. Find the respective probabilities for these combinations: P(BR)=P(RB)=672P(BR)=P(RB)=\frac{6}{72}, P(BG)=P(GB)=1272P(BG)=P(GB)=\frac{12}{72}, P(RG)=P(GR)=872P(RG)=P(GR)=\frac{8}{72}. The probability is: p=12+24+1672=5272=1318p=\frac{12+24+16}{72}=\frac{52}{72}=\frac{13}{18}

The answer is: 1318.\frac{13}{18}.


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