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Anita505

  • 3 years ago

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.

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  1. goformit100
    • 3 years ago
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    I know the solution...

  2. Anita505
    • 3 years ago
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    Can you please help me?

  3. kropot72
    • 3 years ago
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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?

  4. Anita505
    • 3 years ago
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    Sorry i can not calculate it,

  5. Anita505
    • 3 years ago
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    my calculator is currently not working

  6. Anita505
    • 3 years ago
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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)=

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

  8. kropot72
    • 3 years ago
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    Hint: \[\frac{5}{10}\times \frac{5}{10}=\frac{1}{2}\times \frac{1}{2}\]

  9. Anita505
    • 3 years ago
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    so the answer would be 0.75

  10. kropot72
    • 3 years ago
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    Good work! Your answer is correct.

  11. Anita505
    • 3 years ago
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    oops no sorry my mistake

  12. kropot72
    • 3 years ago
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    0.75 or 3/4 is correct.

  13. Anita505
    • 3 years ago
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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?

  14. teddyt477
    • 3 years ago
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    take the chance after by subtracting out already drawn balls

  15. Anita505
    • 3 years ago
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    hmm?

  16. kropot72
    • 3 years ago
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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) \[P(at\ least\ one\ red)=(\frac{1}{2}\times \frac{4}{9})+(\frac{1}{2}\times \frac{5}{9})+(\frac{1}{2}\times \frac{5}{9})=?\]

  17. Anita505
    • 3 years ago
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    thank you and the answer i have recieved was 0.77777777777

  18. Anita505
    • 3 years ago
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    is that also what you got?

  19. kropot72
    • 3 years ago
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    Your answer to B) is correct. Good work :)

  20. Anita505
    • 3 years ago
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    thank you for your help! :)

  21. Anita505
    • 3 years ago
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    but should i put my answer as 0.78?

  22. Anita505
    • 3 years ago
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    it says to simply my answer into a integer or fraction

  23. Anita505
    • 3 years ago
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    simplify**

  24. Anita505
    • 3 years ago
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    oops nvm its 7/9 hahaha thanks!

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

  26. kropot72
    • 3 years ago
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    Yes, 7/9 is correct!

  27. Anita505
    • 3 years ago
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    :)

  28. kropot72
    • 3 years ago
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    You're welcome :)

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