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Anita505

  • one year ago

a bag contains 9 blue marbles, 8 white marbles and 3 red marbles. Three marbles are selected from the bag. What is the probability they are all red? What is the probability that they are all the same colours?

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  1. tomo
    • one year ago
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    P(3 reds) = C(3,3)/C(20,3)

  2. Anita505
    • one year ago
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    how do u solve that please?

  3. johnny0929
    • one year ago
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    at first, the probability of selecting a red marble is \(\frac{3}{20}\), and then it is \(\frac{2}{19}\), then it's \(\frac{1}{18}\). multiply them all to get the probability of selecting all 3 red ones!

  4. Anita505
    • one year ago
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    so multiply all three fractions?

  5. agent0smith
    • one year ago
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    ^ yes multiply all three together \[\frac{ 3 }{ 20 } \times \frac{ 2 }{ 19 } \times \frac{ 1 }{18 } = ???\]

  6. Anita505
    • one year ago
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    to get answer part a?

  7. Anita505
    • one year ago
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    whats for part b?

  8. agent0smith
    • one year ago
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    For probability that they're all the same colours, add up three separate probabilities - prob of getting three white, and prob of getting three blue, and prob of getting three red (which we just found)

  9. Anita505
    • one year ago
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    0.00087719298

  10. johnny0929
    • one year ago
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    @agent0smith and then do you add them together? i forgot how to do that part.

  11. agent0smith
    • one year ago
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    Yeah i think you just add them together

  12. Anita505
    • one year ago
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    that is the answer i just found for multiplying all three?? which answer does that belong to there are two questions, first one is: What is the probability they are all red? and the second question is : What is the probability that they are all the same colours?

  13. johnny0929
    • one year ago
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    the answer you just found was the reds... please read through the explanation agent0smith and I just provided

  14. tomo
    • one year ago
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    more help: https://www.khanacademy.org/math/trigonometry/prob_comb/basic_prob_precalc/v/simple-probability

  15. Anita505
    • one year ago
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    the first part of the question the answer that i got was 0.00087719298..

  16. Anita505
    • one year ago
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    so for part a my answer would be 0.001 and for part b my answer would be 0.31

  17. agent0smith
    • one year ago
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    Show us how you did part b, so someone can check it.

  18. tomo
    • one year ago
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    part b. this is how you do it. C(n,r) is a combination. P(all the same color) = [(C(3,1)*C(9,3)+C(2,1)*C(8,3)+C(1,1)*C(3,3)]/C(20,3) the idea is C(3,1) is choosing the first color. then you have to choose three from that color. I chose blue to start with C(9,3). then you have C(2,1) which is because you have removed one color from the possible colors. then you have to choose 3 from that color. I chose white C(8,3). then you only have one color left C(1,1). this color must be red C(3,3). Then you divide by all possible combinations C(20,3). done.

  19. agent0smith
    • one year ago
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    @tomo are you sure that's correct? Because you'll end up with different probabilities depending on the order you enter them... eg if you choose white first instead of blue.

  20. agent0smith
    • one year ago
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    And a 32% chance of getting all three the same colour seems kinda high. The probability of JUST pulling out one blue from the bag is only 9/20 = 45%. Chance of getting three blues = 7.4%, chance of getting three reds = 0.09%, chance of getting three white = 4.91% Also note that my method and yours get the same probability IF you remove the C(3,1), C(2,1) and C(1,1) from your equation. my method on the left, yours on the right: \[\frac{ (3 \times 2 \times 1) + (9 \times 8 \times 7) + (8 \times 7 \times 6) }{ 20 \times 19 \times 18 } = \frac{ C(9,3) + C(8,3) +C(3,3) }{ C(20,3) }\]

  21. agent0smith
    • one year ago
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    They simplify to \[\frac{ 846 }{ 6840 } = \frac{ 141 }{ 1140 }\] So the prob. of picking three the same colour = 0.124 = 12.4%

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