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

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- anonymous

What have you tried?

- anonymous

Please somebody give the probability of the each of the following.

- dumbsearch2

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- anonymous

Ok @dumbsearch2 pls solve my question

- dumbsearch2

except me. I'm a wannabe-authority.

- dumbsearch2

hey, make a new question. close this one. its too old to get bumped.

- anonymous

I have tried 1/10,2/5,3/10 and 2/5

- terenzreignz

okay... in any month, the odds of getting the flu is 1/10, right?

- terenzreignz

Getting the flu in one month (most unnaturally) does not affect the odds of getting the flu in any other month.

- terenzreignz

Hence, they are independent events, and you can multiply the probabilities as needed.

- anonymous

I did'nt understand @terenzreignz

- terenzreignz

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

- anonymous

Ok @terenzreignz but then how do I do the remaining 3 ???

- terenzreignz

I'll help in the second, ok?

- anonymous

Ok

- terenzreignz

ok, getting the flu in sept and nov, and NOT getting the flu in oct
what are the odds of not getting flu ?

- anonymous

@terenzreignz there no odds of not getting flu

- kropot72

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
\[P(3\ months\ out\ of\ 12)=12C3\times (0.1)^{3}\times (0.9)^{9}=\frac{17}{200}=0.085\]

- kropot72

@Syamantak Are you there?

- anonymous

@kropot72 yes I'm there

- kropot72

Do you understand the method?

- anonymous

Yes,but the answer for the first question would be 1/1000

- anonymous

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

- kropot72

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?

- anonymous

No I can't understand and do part 2,3 and 4.@kropot72

- kropot72

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.

- anonymous

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.

- kropot72

If the correct answer for question 4) is 7/250, then 27/1000 is correct for question 2).

- anonymous

But its coming wrong

- anonymous

Question 3 answer, 0.027, is off by 1/3

- kropot72

(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

Nice explanation. I am math challenged but i get it. Good job.

- kropot72

You're welcome :)

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