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bealiberty47

  • one year ago

Probability math question.

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  1. bealiberty47
    • one year ago
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  2. anonymous
    • one year ago
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    Some questions for you: How many marbles are there in total? How many of them are red? blue?

  3. bealiberty47
    • one year ago
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    16 marbles in total 4 red 7 blue

  4. anonymous
    • one year ago
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    Alright, so the probability of drawing a red is \(\dfrac{\text{number of reds}}{\text{total number of marbles}}\). Similarly, the probability of drawing a blue is \(\dfrac{\text{number of blues}}{\text{total number of marbles}}\)

  5. bealiberty47
    • one year ago
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    Red probability is 4/16 simplified to 1/4 Blue probability is 7/16

  6. bealiberty47
    • one year ago
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    right?

  7. anonymous
    • one year ago
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    Yes. So in the first problem, you're drawing a red marble first, which you know has probability 1/4. When you put the marble back in, you still have 16 total marbles, so the probability of drawing a blue is 7/16. The fact that you replace the first marble means that your draws are independent. This means \[P(\text{red then blue})=P(\text{red})\times P(\text{blue})\]

  8. bealiberty47
    • one year ago
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    so you multiply red, 4, by blue, 7. or you multiply the probabilities of 1/4 and 7/16

  9. anonymous
    • one year ago
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    You multiply the probabilities. \(P(\text{red})=\dfrac{1}{4}\) and \(P(\text{blue})=\dfrac{7}{16}\), so \(P(\text{red then blue})=\dfrac{1}{4}\times\dfrac{7}{16}\).

  10. bealiberty47
    • one year ago
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    sorry was afk my mom called. I'll get right to that :)

  11. bealiberty47
    • one year ago
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    7/64?

  12. anonymous
    • one year ago
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    Right. For the second problem, you're not putting back the first marble. So on the first draw, the probability of drawing a red stays the same, \(\dfrac{1}{4}\). On the second draw, there are now 15 marbles in total, so what's the probability of drawing a blue from these 15?

  13. bealiberty47
    • one year ago
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    7/15

  14. anonymous
    • one year ago
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    Right. As before, the probabilities are independent, so \[P(\text{red, then blue (after replacement)}=P(\text{red})\times P(\text{blue})_{\text{(after replacement)}}\]

  15. bealiberty47
    • one year ago
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    so 1/4 x 7/15 = 7/60

  16. anonymous
    • one year ago
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    Yes

  17. anonymous
    • one year ago
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    The answer to part (c) should be obvious. In the first scenario, you're keeping the number of total marbles the same, but in the second it's being changed.

  18. bealiberty47
    • one year ago
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    yeah ok. I get it. I was pretty much speeding through the lesson cause its the end of school but I do enjoy math when I slow down and understand it. Thanks so much :)

  19. anonymous
    • one year ago
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    You're welcome!

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