## 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

1. drawar

What have you tried?

2. Syamantak

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

3. dumbsearch2

Welcome to OpenStudy :) The people in purple are moderators :)

4. Syamantak

Ok @dumbsearch2 pls solve my question

5. dumbsearch2

except me. I'm a wannabe-authority.

6. dumbsearch2

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

7. Syamantak

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

8. terenzreignz

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

9. terenzreignz

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

10. terenzreignz

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

11. Syamantak

I did'nt understand @terenzreignz

12. 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

13. Syamantak

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

14. terenzreignz

I'll help in the second, ok?

15. Syamantak

Ok

16. terenzreignz

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

@terenzreignz there no odds of not getting flu

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

19. kropot72

@Syamantak Are you there?

20. Syamantak

@kropot72 yes I'm there

21. kropot72

Do you understand the method?

22. Syamantak

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

23. Syamantak

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

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

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

26. 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.

27. Syamantak

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

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

29. Syamantak

But its coming wrong

30. diek

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

31. 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: $P(once\ only)=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\times\frac{1}{10}\times (\frac{9}{10})^{2}=\frac{3}{1}\times\frac{1}{10}\times\frac{81}{100}=\frac{243}{1000}$

32. diek

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

33. kropot72

You're welcome :)