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dboyette

  • 5 years ago

In testing a new drug, researchers found that 20% of all patients using it will have a mild side effect. A random sample of 14 patients using the drug is selected. Find the probability that: (A) exactly four will have this mild side effect (B) at least five will have this mild side effect. Help, I am failing this class

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    exactly 4: 4 will have it, ten will not so your probability is \[P(x=4)=\dbinom{14}{4}.2^4\times .8^{10}\]

  3. anonymous
    • 5 years ago
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    where \[\dbinom{12}{4}\] is the number of ways you can choose 4 people from a set of 14. \[\dbinom{14}{4}=\frac{14\times 13\times 12\times 11}{4\times 3\times 2}=7\times 13\times 11=1001\]

  4. anonymous
    • 5 years ago
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    you need a calculator to continue

  5. anonymous
    • 5 years ago
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    i get .172 rounded.

  6. anonymous
    • 5 years ago
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    "at least 5" is a real pain because you either have to compute the probabilities for X = 0, 1, 2, 3, and 4 and then subtract them from 1.

  7. anonymous
    • 5 years ago
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    i will write them if you like

  8. dboyette
    • 5 years ago
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    please show me, that is where I get lost

  9. anonymous
    • 5 years ago
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    ok. we put X = number of cases of side effects. you are given that the probabiltiy any one person has it is 20%=.2 so the probability any one person does not have it is 80% = .8

  10. anonymous
    • 5 years ago
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    easy to compute the probability no one has it, because it is just \[.8^{14}\]

  11. anonymous
    • 5 years ago
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    in other words you get 14 in a row without the side effect. is that much clear?

  12. dboyette
    • 5 years ago
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    yes it is thanks

  13. anonymous
    • 5 years ago
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    ok now let us compute the probability that one person has the side effect. one person has it, that means 13 do not. so it is going to look like \[.2^1\times .8^{13}\]

  14. anonymous
    • 5 years ago
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    but there are 14 different ways for this to happen. either the first person has the side effect, or the second , or the third , or so on. so the probabily that on person has it is \[P(X=1)=14\times .2^1\times .8^{13}\]

  15. dboyette
    • 5 years ago
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    so P(X=2)= 14*.2^2*.8^12?

  16. anonymous
    • 5 years ago
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    hello armistre

  17. amistre64
    • 5 years ago
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    howdy ;)

  18. anonymous
    • 5 years ago
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    bdoyette you are close. one part will be \[.2^2\times .8^{12}\]

  19. anonymous
    • 5 years ago
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    but this time you don't multiply by 14, you multiply by the number of ways you can choose 2 people out of 14

  20. anonymous
    • 5 years ago
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    this is usually written as \[\dbinom{14}{2}\] and is computed as follows: \[\dbinom{14}{2}=\frac{14\times 13}{2}=7\times 13=91\]

  21. anonymous
    • 5 years ago
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    so \[P(X=2)=91\times .2^2\times .8^{12}\]

  22. anonymous
    • 5 years ago
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    i will do the next one too if you like

  23. dboyette
    • 5 years ago
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    so the next one would be 14/3=14*13*12/3?

  24. anonymous
    • 5 years ago
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    \[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}\]

  25. anonymous
    • 5 years ago
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    think as follows. I pick three from a set of 14. i have 14 choices for the first, 13 for the second and 12 for the third. this explains the numerator. but since i don't care what order i pick them in i have counted too many ways. the numerator counts (ABC) different than (ACB) different than (BAC) etc. so i have to divide by how many ways i can permute these three people which is 3*2=6

  26. anonymous
    • 5 years ago
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    another example would be \[\dbinom{14}{5}=\frac{14\times 13\times 12\times 11\times 10\times 9}{5\times 4\times 3\times 2}\]

  27. dboyette
    • 5 years ago
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    so that would be 364* 0.2^3*0.8^11?

  28. dboyette
    • 5 years ago
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    the answer for 3

  29. anonymous
    • 5 years ago
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    yes i think i didn't compute it so let me check

  30. anonymous
    • 5 years ago
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    yes.

  31. dboyette
    • 5 years ago
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    the # 5 would be 18018*0.2^5*0.8^9?

  32. anonymous
    • 5 years ago
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    \[\dbinom{14}{3}=\frac{14\times 13\times 12}{3\times 2}=14\times 13\times 2=364\]

  33. anonymous
    • 5 years ago
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    yes you have it

  34. dboyette
    • 5 years ago
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    omg i think i am finally starting to get this, thank you so much

  35. anonymous
    • 5 years ago
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    i can write a general formula for you if you like

  36. dboyette
    • 5 years ago
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    please

  37. anonymous
    • 5 years ago
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    ok you in this example you have a probability that is 20%=.2 in the general case you would have a probability that is a variable, call it p

  38. anonymous
    • 5 years ago
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    we knew that if the probability of having a side effect was .2 then the probability of not having it was .8 because they have to add to 1

  39. dboyette
    • 5 years ago
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    ok

  40. anonymous
    • 5 years ago
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    so in the general case we would have the probability of having it is p, and the probability of not having it is 1-p

  41. anonymous
    • 5 years ago
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    in this case we had 14 people, in the general case we will have n people (tosses of a coin, whatever)

  42. anonymous
    • 5 years ago
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    so if i want to know the probability that 5 people have the side effect, that means that 9 don't. in the general case if i want to know the probability that k people have it that means that n-k will not. so far so good?

  43. dboyette
    • 5 years ago
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    yes

  44. anonymous
    • 5 years ago
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    ok now we computed the probability that 5 people had it by computing \[.2^{5} \times .8^9\] and then multiplying by the number of ways to select 5 people from 9. in the general case we will have k people with the side effect and n - k without, so you will have to compute \[p^k\times (1-p)^{n-k}\]

  45. anonymous
    • 5 years ago
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    that is like our \[.2^5\times .8^9\]

  46. anonymous
    • 5 years ago
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    but then you will have to multiply by the number of ways you can choose those 5 out of 14, or in this case the number of ways you can choose k out of n. that is written as \[\dbinom{n}{k}\]

  47. anonymous
    • 5 years ago
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    so the "final answer" is, if you have n independent trials with probability of success = p and therefore the probability of failure = 1-p at each trial (these are called bernoulli trials) the the probability you get exactly k success is:... \[P(X=k)=\dbinom{n}{k}p^k(1-p)^{n-k}\]

  48. anonymous
    • 5 years ago
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    we computed this with n = 14, p = .2 and 1-p=.8 for various different values of k

  49. dboyette
    • 5 years ago
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    so k was the different numbers we used for side effects

  50. anonymous
    • 5 years ago
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    yes, the number of people who could have it. question one was for k = 4

  51. anonymous
    • 5 years ago
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    to do question 2 you have to compute the probability \[P(X=0)\] which we did, the probability \[P(X=1)\] which we also did, the probability \[P(X=2)\] \[P(X=3)\] \[P(X=4)\] a real pain. then we add these numbers up and subtract them from 1

  52. anonymous
    • 5 years ago
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    because "5 or more" means "not none, not one, not two, not three and not four"

  53. dboyette
    • 5 years ago
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    what if they don't equal 1?

  54. anonymous
    • 5 years ago
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    now sure what you are asking.

  55. anonymous
    • 5 years ago
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    the total is 1 for all the possibilities. it has to be, because the probability of the whole sample space is always one.

  56. anonymous
    • 5 years ago
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    just like if i know that the probability i roll a 10 on two dice is \[\frac{4}{36}\] then i know the probability i don't is \[\frac{32}{36}\]

  57. dboyette
    • 5 years ago
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    #2 91*.2^2*.8^12 ==0.25 #3 364*0.2^3*0.8^11 = 0.25 #4 1001*0.2^4*0.8^10 = 0.17 5# 18018*0.2^5*0.8^9=0.77 these don't add up to 1

  58. anonymous
    • 5 years ago
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    ok let me go slow. the total is one, meaning that if you compute all the probabilities for x = 0, x =1, x = 2, x= 3, x = 4, x = 5, x = 6, x = 7, x = 8, x = 9, x = 10, x = 11, x = 12, x = 13, x = 14 and add them up you will get 1

  59. anonymous
    • 5 years ago
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    you only want at least 5, meaning x = 5, x = 6, x = 7, x = 8, x = 9, x = 10 , x = 11, x = 12, x = 13, x = 14

  60. anonymous
    • 5 years ago
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    you can compute all these if you want, and add them and that will be your answer.

  61. anonymous
    • 5 years ago
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    but it is easier to compute the probabilities for x = 0, x = 1, x = 2, x = 3, and x = 4. add them up and then subtract this number from 1

  62. anonymous
    • 5 years ago
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    this way you only have to compute 5 probabilities instead of 10

  63. dboyette
    • 5 years ago
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    oh i see. that makes sense

  64. anonymous
    • 5 years ago
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    good!

  65. anonymous
    • 5 years ago
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    btw you do not need to compute \[P(X=5)\] because it says "at least 5" so you compute just the lower ones and subtract the result from 1

  66. dboyette
    • 5 years ago
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    thank you you are awesome!

  67. anonymous
    • 5 years ago
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    welcome!

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