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Anita505 Group Title

Two balls are drawn in succession out of a box containing 5 red and 5 white balls. Find the probability that at least 1 ball was red, given that the first ball was.. A) Replaced before the second draw B) NOT replaced before the second draw a) Find the probability that at least 1 ball was red, given that the first ball was replaced before the second draw.

  • one year ago
  • one year ago

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  1. goformit100 Group Title
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    I know the solution...

    • one year ago
  2. Anita505 Group Title
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    Can you please help me?

    • one year ago
  3. kropot72 Group Title
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    Section A) Consider the following 3 situations: (1) A: First ball red B: Second ball red The events A and B are independent therefore P(two red) = (5/10) * (5/10) ...........................(1) (2) A: First ball red B: Second ball black The events A and B are independent therefore P(one red) = (5/10) * (5/10) ...........................(2) (3) A: First ball black B: Second ball red The events A and B are independent therefore P(one red) = (5/10) * (5/10) ...........................(3) The probability that at least 1 ball was red, given that the first ball was replaced before the second draw is the sum of results (1), (2) and (3). Can you calculate?

    • one year ago
  4. Anita505 Group Title
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    Sorry i can not calculate it,

    • one year ago
  5. Anita505 Group Title
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    my calculator is currently not working

    • one year ago
  6. Anita505 Group Title
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    so this is what it would look like then (5/10)(5/10)+(5/10)(5/10)+(5/10)(5/10)=

    • one year ago
  7. kropot72 Group Title
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    You do not need a calculator to work this out: \[P(at\ least\ one\ red)=(\frac{5}{10}\times \frac{5}{10})+(\frac{5}{10}\times \frac{5}{10})+(\frac{5}{10}\times \frac{5}{10})=?\]

    • one year ago
  8. kropot72 Group Title
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    Hint: \[\frac{5}{10}\times \frac{5}{10}=\frac{1}{2}\times \frac{1}{2}\]

    • one year ago
  9. Anita505 Group Title
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    so the answer would be 0.75

    • one year ago
  10. kropot72 Group Title
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    Good work! Your answer is correct.

    • one year ago
  11. Anita505 Group Title
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    oops no sorry my mistake

    • one year ago
  12. kropot72 Group Title
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    0.75 or 3/4 is correct.

    • one year ago
  13. Anita505 Group Title
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    How would i find the probability that at least 1 ball was red, given that the first ball s not replaced before the second draw?

    • one year ago
  14. teddyt477 Group Title
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    take the chance after by subtracting out already drawn balls

    • one year ago
  15. Anita505 Group Title
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    hmm?

    • one year ago
  16. kropot72 Group Title
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    Section B) Probability first ball is red = 5/10 Probability second ball is red = 4/9 P(two red) = (5/10 * 4/9) ................(1) Probability first ball is red = 5/10 Probability second ball is black = 5/9 P(one red) = (5/10 * 5/9) ...............(2) Probability first ball is black = 5/10 Probability second ball is red = 5/9 P(one red) = (5/10 * 5/9) ...............(3) \[P(at\ least\ one\ red)=(\frac{1}{2}\times \frac{4}{9})+(\frac{1}{2}\times \frac{5}{9})+(\frac{1}{2}\times \frac{5}{9})=?\]

    • one year ago
  17. Anita505 Group Title
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    thank you and the answer i have recieved was 0.77777777777

    • one year ago
  18. Anita505 Group Title
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    is that also what you got?

    • one year ago
  19. kropot72 Group Title
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    Your answer to B) is correct. Good work :)

    • one year ago
  20. Anita505 Group Title
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    thank you for your help! :)

    • one year ago
  21. Anita505 Group Title
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    but should i put my answer as 0.78?

    • one year ago
  22. Anita505 Group Title
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    it says to simply my answer into a integer or fraction

    • one year ago
  23. Anita505 Group Title
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    simplify**

    • one year ago
  24. Anita505 Group Title
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    oops nvm its 7/9 hahaha thanks!

    • one year ago
  25. kropot72 Group Title
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    To put the answer as a fraction just add the following fractions and simplify: \[\frac{4}{18}+\frac{5}{18}+\frac{5}{18}=?\]

    • one year ago
  26. kropot72 Group Title
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    Yes, 7/9 is correct!

    • one year ago
  27. Anita505 Group Title
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    :)

    • one year ago
  28. kropot72 Group Title
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    You're welcome :)

    • one year ago
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