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Syamantak Group Title

Let's say Alvin will catch the flu with probability of 1/10 during any given month.Let's also assume that Alvin can catch the flu only once per month, and that if he has caught the flu,the flu virus will die by the end of the month.What is the probability of the following events? 1)He catches the flu in September, October and November. 2)He catches the flu in September and then again in November, but not in October. 3)He catches the flu exactly once in the three months from September through November. 4)He catches the flu in two or more of the three months from September through November

  • one year ago
  • one year ago

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  1. drawar Group Title
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    What have you tried?

    • one year ago
  2. Syamantak Group Title
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    Please somebody give the probability of the each of the following.

    • one year ago
  3. dumbsearch2 Group Title
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    Welcome to OpenStudy :) The people in purple are moderators :)

    • one year ago
  4. Syamantak Group Title
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    Ok @dumbsearch2 pls solve my question

    • one year ago
  5. dumbsearch2 Group Title
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    except me. I'm a wannabe-authority.

    • one year ago
  6. dumbsearch2 Group Title
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    hey, make a new question. close this one. its too old to get bumped.

    • one year ago
  7. Syamantak Group Title
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    I have tried 1/10,2/5,3/10 and 2/5

    • one year ago
  8. terenzreignz Group Title
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    okay... in any month, the odds of getting the flu is 1/10, right?

    • one year ago
  9. terenzreignz Group Title
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    Getting the flu in one month (most unnaturally) does not affect the odds of getting the flu in any other month.

    • one year ago
  10. terenzreignz Group Title
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    Hence, they are independent events, and you can multiply the probabilities as needed.

    • one year ago
  11. Syamantak Group Title
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    I did'nt understand @terenzreignz

    • one year ago
  12. terenzreignz Group Title
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    say the first question... sept, oct, and nov the odds of getting the flu in each of those months is 1/10 each so just multiply sept = 1/10 oct = 1/10 nov = 1/10 multiply them 1/1000

    • one year ago
  13. Syamantak Group Title
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    Ok @terenzreignz but then how do I do the remaining 3 ???

    • one year ago
  14. terenzreignz Group Title
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    I'll help in the second, ok?

    • one year ago
  15. Syamantak Group Title
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    Ok

    • one year ago
  16. terenzreignz Group Title
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    ok, getting the flu in sept and nov, and NOT getting the flu in oct what are the odds of not getting flu ?

    • one year ago
  17. Syamantak Group Title
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    @terenzreignz there no odds of not getting flu

    • one year ago
  18. kropot72 Group Title
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    The binomial distribution can be used to solve this one: 1) The probability of catching the flu in 3 months out of 12 is given by \[P(3\ months\ out\ of\ 12)=12C3\times (0.1)^{3}\times (0.9)^{9}=\frac{17}{200}=0.085\]

    • one year ago
  19. kropot72 Group Title
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    @Syamantak Are you there?

    • one year ago
  20. Syamantak Group Title
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    @kropot72 yes I'm there

    • one year ago
  21. kropot72 Group Title
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    Do you understand the method?

    • one year ago
  22. Syamantak Group Title
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    Yes,but the answer for the first question would be 1/1000

    • one year ago
  23. Syamantak Group Title
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    As the probability of catching flu in one month is 1/10.So the probability of catching it in 3 months would be 1/10 * 1/10 * 1/10 = 1/1000

    • one year ago
  24. kropot72 Group Title
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    My answer is correct for the probability of catching the flu 3 months out of 12 months and not catching the flu in any of the other 9 months. If the question relates to only the 3 months that are named then 1/1000 is correct. Can you do the other parts of the question?

    • one year ago
  25. Syamantak Group Title
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    No I can't understand and do part 2,3 and 4.@kropot72

    • one year ago
  26. kropot72 Group Title
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    3) The probability of catching the flu in 1 month out of 3 months is \[P(flu\ 1\ month\ out\ of\ 3)=3C1\times 0.1\times (0.9)^{2}=3\times 0.1\times (0.9)^{2}=you\ can\ calculate\] 4) Add the results of 1) and 2) to find the probability he catches the flu in two or more of the three months from September through November.

    • one year ago
  27. Syamantak Group Title
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    I got all the answers @kropot72 but the answer you gave me for question 2 was wrong.All the other answers are correct and the answer for question 4 will be 7/250.

    • one year ago
  28. kropot72 Group Title
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    If the correct answer for question 4) is 7/250, then 27/1000 is correct for question 2).

    • one year ago
  29. Syamantak Group Title
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    But its coming wrong

    • one year ago
  30. diek Group Title
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    Question 3 answer, 0.027, is off by 1/3

    • 7 months ago
  31. kropot72 Group Title
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    (2) The probability of catching the flu in September and November but not in October is: \[\frac{1}{10}\times\frac{9}{10}\times\frac{1}{10}=\frac{9}{1000}\] (3) The probability of catching the flu exactly once in the three months is: \[P(once\ only)=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\times\frac{1}{10}\times (\frac{9}{10})^{2}=\frac{3}{1}\times\frac{1}{10}\times\frac{81}{100}=\frac{243}{1000}\]

    • 7 months ago
  32. diek Group Title
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    Nice explanation. I am math challenged but i get it. Good job.

    • 7 months ago
  33. kropot72 Group Title
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    You're welcome :)

    • 7 months ago
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