## 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

1. drawar Group Title

What have you tried?

2. Syamantak Group Title

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

3. dumbsearch2 Group Title

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

4. Syamantak Group Title

Ok @dumbsearch2 pls solve my question

5. dumbsearch2 Group Title

except me. I'm a wannabe-authority.

6. dumbsearch2 Group Title

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

7. Syamantak Group Title

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

8. terenzreignz Group Title

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

9. terenzreignz Group Title

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

10. terenzreignz Group Title

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

11. Syamantak Group Title

I did'nt understand @terenzreignz

12. terenzreignz Group Title

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 Group Title

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

14. terenzreignz Group Title

I'll help in the second, ok?

15. Syamantak Group Title

Ok

16. terenzreignz Group Title

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 Group Title

@terenzreignz there no odds of not getting flu

18. kropot72 Group Title

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 Group Title

@Syamantak Are you there?

20. Syamantak Group Title

@kropot72 yes I'm there

21. kropot72 Group Title

Do you understand the method?

22. Syamantak Group Title

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

23. Syamantak Group Title

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 Group Title

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 Group Title

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

26. kropot72 Group Title

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 Group Title

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 Group Title

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

29. Syamantak Group Title

But its coming wrong

30. diek Group Title

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

31. kropot72 Group Title

(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 Group Title

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

33. kropot72 Group Title

You're welcome :)