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anonymous

  • one year ago

You are choosing two beads there are 40 beads. 7/40 are yellow 5/40 are green 9/40 are blue 8/40 are brown 7/40 are orange 4/40 are red calculate WITH replacement and WITHOUT replacement for each: P(R1B2 OR B1R2) (R1 refers to chosing red the first time and B2 refers to picking Brown the second time)

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  1. ash2326
    • one year ago
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    @butterflyprincess did you start working on the problem?

  2. anonymous
    • one year ago
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    yea I gt 99/2500 for with replacement but I want to know if it is correct

  3. ash2326
    • one year ago
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    ok, let me calculate and see if our answer matches

  4. anonymous
    • one year ago
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    okay

  5. ash2326
    • one year ago
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    My answer for with replacement is not matching with yours. Could you describe here how did you solve the problem?

  6. anonymous
    • one year ago
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    oh yea so I used P(A or B)=P(A)+P(B)-P(AandB) so I got like (32/1600)+(32/1600)-(1024/2560000)

  7. ash2326
    • one year ago
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    How A and B can occur at the same time?

  8. anonymous
    • one year ago
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    oh no

  9. ash2326
    • one year ago
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    P (A and B)= 0, so I think with replacement is solved now.

  10. welshfella
    • one year ago
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    Hint: R1B2 = red first AND blue second AND = multiply probabilities OR = add them

  11. anonymous
    • one year ago
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    wait so would it be then (32/1600)+(32/1600)

  12. ash2326
    • one year ago
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    yep,

  13. anonymous
    • one year ago
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    and then for without replacement it will be 32/1560 + 32/1560

  14. anonymous
    • one year ago
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    because (4/40)(8/39)+(8/40)(4/39)

  15. welshfella
    • one year ago
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    yes

  16. ash2326
    • one year ago
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    yes, @butterflyprincess you got it right.

  17. anonymous
    • one year ago
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    wait so if they ask P(R1 and G2) then it would be 0 since both events can't happen at the same time/?

  18. ash2326
    • one year ago
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    P (R1 and G2)= probability of red bead first and green beard second is a valid event and can occur

  19. anonymous
    • one year ago
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    which is 20/1600 (repl) and 20/1560 (without repl) right

  20. ash2326
    • one year ago
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    but if they say P(R1 and G1) =0 since both red and green can't be first at the same time.

  21. welshfella
    • one year ago
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    - but they cant occur together on one pick , of course

  22. anonymous
    • one year ago
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    makes sense so that would be 20/1600 (repl) and 20/1560 (without repl) right?

  23. ash2326
    • one year ago
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    correct :)

  24. anonymous
    • one year ago
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    nice cool wait sorry I have another question lol so I also need to find P(No Yellows) P(doubles) and P(no doubles)

  25. anonymous
    • one year ago
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    for no yellows would it be 1089/1600 (with repl) and 1056/1560 (without)

  26. ash2326
    • one year ago
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    yes, no yellows answer is right.

  27. anonymous
    • one year ago
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    okay and how can I approach the doubles and no double problem

  28. ash2326
    • one year ago
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    doubles mean.. same color both time. So on both draws we should get same color

  29. ash2326
    • one year ago
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    either both yellows, or green or brown or blue or red or orange. find probability for each case and add them all P (Y1 Y2)+ P(G1 G2)+P(Br 1 Br 2)+P (BL1+BL2)+ P(R1 R2)+ P(O1 O2)

  30. anonymous
    • one year ago
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    cant you do like (1/6)(1/6) then without is (1/6)(1/5)

  31. ash2326
    • one year ago
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    If we had only one bead for each color, we could do that. But the no. of each color beads is more than 1, we can't apply that.

  32. ash2326
    • one year ago
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    do you understand?

  33. anonymous
    • one year ago
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    yea oh okay so it would be like (7/40)(7/40)) +(5/40)(5/40)......

  34. ash2326
    • one year ago
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    right.. with replacement.

  35. anonymous
    • one year ago
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    so would be like 0.1775 right right then without replacement would be like (7/40)(6/39)+(5/40)(4/39).....

  36. ash2326
    • one year ago
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    yes, you are right.

  37. welshfella
    • one year ago
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    probabil-ty of no doubles will i think be = 1 - p(doubles) right?

  38. anonymous
    • one year ago
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    right! So like 1-0.1775!!

  39. ash2326
    • one year ago
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    yes, p (non doubles)= 1 -p (doubles)

  40. anonymous
    • one year ago
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    cool thanks for your help! oh one more so P(G2 | R1) would be 5/40 (with repl) and 5/39 withut

  41. ash2326
    • one year ago
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    Green in second provided Red is drawn first. Yes you are right.

  42. anonymous
    • one year ago
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    nice thanks

  43. ash2326
    • one year ago
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    no problem :)

  44. ash2326
    • one year ago
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    You can close the question, if you have no more parts of this question. Ask a new one if you need help with a different problem. Thanks!!

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