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Syamantak

  • one year ago

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

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  1. drawar
    • one year ago
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    What have you tried?

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

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

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

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

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

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

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

  9. terenzreignz
    • one year ago
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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.

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

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

  12. terenzreignz
    • one year ago
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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

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

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

  15. Syamantak
    • one year ago
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    Ok

  16. terenzreignz
    • one year ago
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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 ?

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

  18. kropot72
    • one year ago
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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 \[P(3\ months\ out\ of\ 12)=12C3\times (0.1)^{3}\times (0.9)^{9}=\frac{17}{200}=0.085\]

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

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

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

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

  23. Syamantak
    • one year ago
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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

  24. kropot72
    • one year ago
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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?

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

  26. kropot72
    • one year ago
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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.

  27. Syamantak
    • one year ago
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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.

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

  29. Syamantak
    • one year ago
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    But its coming wrong

  30. diek
    • 8 months ago
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    Question 3 answer, 0.027, is off by 1/3

  31. kropot72
    • 8 months ago
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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: \[P(once\ only)=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\times\frac{1}{10}\times (\frac{9}{10})^{2}=\frac{3}{1}\times\frac{1}{10}\times\frac{81}{100}=\frac{243}{1000}\]

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

  33. kropot72
    • 8 months ago
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    You're welcome :)

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