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briana.img

  • one year ago

Someone help me understand this problem?

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  1. briana.img
    • one year ago
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    A bag has 4 green marbles, 3 red marbles, and 3 yellow marbles. What is the probability that you pick a green marble, do not replace it, and pick a red marble?

  2. Owlcoffee
    • one year ago
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    Well, I believe these are two scenarios. First calculate the probability you have to pick a green marble, you count all the marbles and and the green marbles: \[\frac{ 4 }{ 10 }.100\]

  3. briana.img
    • one year ago
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    why the 100?

  4. Owlcoffee
    • one year ago
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    it's there by mere definition, but it's a formula that gives you the probability.

  5. Owlcoffee
    • one year ago
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    You have to first perform the division, remember that.

  6. briana.img
    • one year ago
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    I got 40

  7. Owlcoffee
    • one year ago
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    Good, now, that gives the probability you would have to get a green marble, now, you won't put it back in the bag, so now you have a total of 9 marbles inside the bag. What would be the probability to find a red marble in the new quatity?: \[\frac{ 3 }{ 9 }.100\]

  8. briana.img
    • one year ago
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    got 33.33

  9. perl
    • one year ago
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    Better not multiply by a 100. Only if you want to convert the final answer to a percentage $$ \Large \sf Probability = \frac {4}{10}\cdot \frac{3}{9}$$

  10. briana.img
    • one year ago
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    2/15??

  11. perl
    • one year ago
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    yes

  12. briana.img
    • one year ago
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    awwwh okay thank you! i'll look at this next time i have trouble with this

  13. Owlcoffee
    • one year ago
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    Haha, I was doing it by parts, but thanks, perl.

  14. briana.img
    • one year ago
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    @perl @Owlcoffee so for the problem A bag has 5 red marbles, 6 blue marbles and 4 black marbles. What is the probability of picking a blue marble, replacing it, and then picking another blue marble? the answer would be 4/25 ?

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