Question

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

Answer 1

What have you tried?

Answer 2

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

Answer 3

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

Answer 4

Ok @dumbsearch2 pls solve my question

Answer 5

except me. I'm a wannabe-authority.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

I did'nt understand @terenzreignz

Answer 12

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

Answer 13

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

Answer 14

I'll help in the second, ok?

Answer 15

Ok

Answer 16

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

Answer 17

@terenzreignz there no odds of not getting flu

Answer 18

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

Answer 19

@Syamantak Are you there?

Answer 20

@kropot72 yes I'm there

Answer 21

Do you understand the method?

Answer 22

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

Answer 23

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

Answer 24

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?

Answer 25

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

Answer 26

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.

Answer 27

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.

Answer 28

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

Answer 29

But its coming wrong

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

You're welcome :)