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anonymous
 3 years 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
anonymous
 3 years 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Please somebody give the probability of the each of the following.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Welcome to OpenStudy :) The people in purple are moderators :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok @dumbsearch2 pls solve my question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0except me. I'm a wannabeauthority.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey, make a new question. close this one. its too old to get bumped.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have tried 1/10,2/5,3/10 and 2/5

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0okay... in any month, the odds of getting the flu is 1/10, right?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Getting the flu in one month (most unnaturally) does not affect the odds of getting the flu in any other month.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Hence, they are independent events, and you can multiply the probabilities as needed.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I did'nt understand @terenzreignz

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0say 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok @terenzreignz but then how do I do the remaining 3 ???

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I'll help in the second, ok?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0ok, getting the flu in sept and nov, and NOT getting the flu in oct what are the odds of not getting flu ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@terenzreignz there no odds of not getting flu

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.3The 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 \[P(3\ months\ out\ of\ 12)=12C3\times (0.1)^{3}\times (0.9)^{9}=\frac{17}{200}=0.085\]

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.3@Syamantak Are you there?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@kropot72 yes I'm there

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.3Do you understand the method?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes,but the answer for the first question would be 1/1000

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0As 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

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.3My 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No I can't understand and do part 2,3 and 4.@kropot72

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.33) 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.3If the correct answer for question 4) is 7/250, then 27/1000 is correct for question 2).

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Question 3 answer, 0.027, is off by 1/3

kropot72
 2 years ago
Best ResponseYou've already chosen the best response.3(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: \[P(once\ only)=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\times\frac{1}{10}\times (\frac{9}{10})^{2}=\frac{3}{1}\times\frac{1}{10}\times\frac{81}{100}=\frac{243}{1000}\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Nice explanation. I am math challenged but i get it. Good job.
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