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anonymous

  • 5 years ago

A fair die is tossed 4 times. Find the probability of a. at least 2 5s b. at most 3 even numbers c. no more than 1 odd number d. at least 1 prime number e. at most 1 number greater than 4

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  1. Yuki
    • 5 years ago
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    you use the fact that P(A) = 1- P(not A)

  2. anonymous
    • 5 years ago
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    can u show me how to do each part?

  3. Yuki
    • 5 years ago
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    let's do a

  4. Yuki
    • 5 years ago
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    P(at least two 5s) = 1- P(not at least two 5s)

  5. Yuki
    • 5 years ago
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    = 1- P(at most one 5)

  6. Yuki
    • 5 years ago
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    so let's find P(at most one 5)

  7. Yuki
    • 5 years ago
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    can you find this ?

  8. anonymous
    • 5 years ago
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    um

  9. anonymous
    • 5 years ago
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    I got 146/1296 for A

  10. anonymous
    • 5 years ago
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    is that right???

  11. Yuki
    • 5 years ago
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    I am not finding the answer, just trying to help

  12. Yuki
    • 5 years ago
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    so P(at most one 5) = P(no 5s) + P(exactly one 5)

  13. Yuki
    • 5 years ago
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    P(no 5) = \[(5/6)^4\]

  14. Yuki
    • 5 years ago
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    P(exactly one 5) follows the binomial distribution, so \[4C1*(1/5)^1*(5/6)^4\]

  15. Yuki
    • 5 years ago
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    when you subtract the sum of those from 1, you get the answer for a),

  16. anonymous
    • 5 years ago
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    wait

  17. Yuki
    • 5 years ago
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    oops, the 1/5 should be 1/6

  18. anonymous
    • 5 years ago
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    don't u need 0 5s as well?

  19. Yuki
    • 5 years ago
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    I already did that, it is (5/6)^4

  20. anonymous
    • 5 years ago
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    wait question is at least 25s

  21. Yuki
    • 5 years ago
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    let me explain you that par again

  22. Yuki
    • 5 years ago
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    P( at least two 5s) = P(exactly two) + P(exactly three)+P(exactly four)

  23. anonymous
    • 5 years ago
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    oh yea I did that

  24. Yuki
    • 5 years ago
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    it is a pain to calculate all those, so we use the rule P(event) = 1-P(complement of that event)

  25. anonymous
    • 5 years ago
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    4C2(1/6)^2(5/6)^3 + 4C3(1/6)^3(5/6)^1 + 4C4(1/6)^4(5/6)^0

  26. anonymous
    • 5 years ago
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    is that right?

  27. Yuki
    • 5 years ago
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    Yes.

  28. anonymous
    • 5 years ago
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    k

  29. anonymous
    • 5 years ago
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    B

  30. Yuki
    • 5 years ago
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    what I did was 1- (5/6)^6 -4C1(1/6)(5/6)^3

  31. Yuki
    • 5 years ago
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    for B, the complement is "no evens at all"

  32. Yuki
    • 5 years ago
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    so 1-P(no evens) will be the answer

  33. anonymous
    • 5 years ago
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    isn't it \[4C3(1/2)^3(1/2)^1 + 4C2(1/2)^2(1/2)^2 + 4C1 (1/2)^1(1/2)^3\]

  34. anonymous
    • 5 years ago
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    ???

  35. Yuki
    • 5 years ago
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    you need to include 4C0(1/2)^4

  36. anonymous
    • 5 years ago
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    oh yea

  37. anonymous
    • 5 years ago
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    everything else is correct?

  38. Yuki
    • 5 years ago
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    az, just to let you know you probably know how to get the answers for all of this. But you also need to know that you are missing the point of what these problems are asking.

  39. Yuki
    • 5 years ago
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    Yes.

  40. anonymous
    • 5 years ago
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    next one idk

  41. anonymous
    • 5 years ago
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    so it can be 0 or 1 odd number?

  42. anonymous
    • 5 years ago
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    ??

  43. Yuki
    • 5 years ago
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    yep

  44. anonymous
    • 5 years ago
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    4C1(1/2)^1 (1/2)^3 + 4C0(1/2)^0 (1/2)^4 ??

  45. anonymous
    • 5 years ago
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    is that right

  46. Yuki
    • 5 years ago
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    yes

  47. Yuki
    • 5 years ago
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    az, I need to go soon so let me leave you an advice

  48. Yuki
    • 5 years ago
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    your method is logically straight forward, and as long as you don't mind calculating a lot of numbers you will get the answer so I have no doubt that you will get any of these wrong from what I have seen so far

  49. anonymous
    • 5 years ago
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    wait

  50. anonymous
    • 5 years ago
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    can we go over the last two really quickly

  51. Yuki
    • 5 years ago
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    but all of the problems here can be found a lot more easily using the fact P(A) = 1-P(notA)

  52. anonymous
    • 5 years ago
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    4C1 (1/2)^1(1/2)^3 + 4C0 (1/2)^0 (1/2)^4 ????

  53. Yuki
    • 5 years ago
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    so d is 1 - P (all of them being prime) and e is P(0) + P(1)

  54. anonymous
    • 5 years ago
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    is that right? what I just put

  55. Yuki
    • 5 years ago
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    nope

  56. anonymous
    • 5 years ago
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    what is it then

  57. Yuki
    • 5 years ago
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    sorry, ask someone else since I need to go. :)

  58. anonymous
    • 5 years ago
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    ????

  59. anonymous
    • 5 years ago
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    what is it then

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