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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
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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drawar Group TitleBest ResponseYou've already chosen the best response.0
What have you tried?
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
Please somebody give the probability of the each of the following.
 one year ago

dumbsearch2 Group TitleBest ResponseYou've already chosen the best response.0
Welcome to OpenStudy :) The people in purple are moderators :)
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
Ok @dumbsearch2 pls solve my question
 one year ago

dumbsearch2 Group TitleBest ResponseYou've already chosen the best response.0
except me. I'm a wannabeauthority.
 one year ago

dumbsearch2 Group TitleBest ResponseYou've already chosen the best response.0
hey, make a new question. close this one. its too old to get bumped.
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
I have tried 1/10,2/5,3/10 and 2/5
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
okay... in any month, the odds of getting the flu is 1/10, right?
 one year ago

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

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
Hence, they are independent events, and you can multiply the probabilities as needed.
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
I did'nt understand @terenzreignz
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
Ok @terenzreignz but then how do I do the remaining 3 ???
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
I'll help in the second, ok?
 one year ago

terenzreignz Group TitleBest ResponseYou've already chosen the best response.0
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

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
@terenzreignz there no odds of not getting flu
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.3
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

kropot72 Group TitleBest ResponseYou've already chosen the best response.3
@Syamantak Are you there?
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
@kropot72 yes I'm there
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.3
Do you understand the method?
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
Yes,but the answer for the first question would be 1/1000
 one year ago

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
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

kropot72 Group TitleBest ResponseYou've already chosen the best response.3
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

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
No I can't understand and do part 2,3 and 4.@kropot72
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.3
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

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
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

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

Syamantak Group TitleBest ResponseYou've already chosen the best response.0
But its coming wrong
 one year ago

diek Group TitleBest ResponseYou've already chosen the best response.0
Question 3 answer, 0.027, is off by 1/3
 6 months ago

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

diek Group TitleBest ResponseYou've already chosen the best response.0
Nice explanation. I am math challenged but i get it. Good job.
 6 months ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.3
You're welcome :)
 6 months ago
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