Probability of the disease people

[Old Student]

Dear helper!

I would like the ask the following questions! Could you help me? Thanks

(a) in city X, the proportion of people having a respiratory disease is 0.08. If ten people in city X are randomly selected, find the probability of that

(i) exactly three people have respiratory disease

(ii) more than two people have respiratory disease

(b) in city Y, 16% of new born babies are under weighted. If a new born baby is under weighted, the probability that its mother being a smoker is 0.58, if a new born baby is not under weight, the probability that its mother being a smoker is 0.27

(i) construct a tree diagram to describe these event

(ii) what is the probability that a new born baby is under weighted if its mother is a non smoker

 

[HallsofIvy]

Help you how? What can you do on these problems and where did you have difficulty? Do you know what the "binomial distribution" is? These look like homework problems. If you have been given these problems in a course, surely you were taught something about how to do them.

 

[Old Student]

I am an old part time student to study statistic now! Even I know what the binomial distribution is, but I really don't know how to solve this question! Just want ask for people's help!

So I can study and know the problem via the procedure!

thank you

 

[HallsofIvy]

If the probability of an event is p (so the probability of it not happening is 1- p) then the probability of it happening m out of n times is \(\displaystyle \frac{n!}{m!(n- m)!}p^n (1- p)^{n- m}\). That is the binomial distribution and is the answer to (ai) with n= 10, m= 3, and p= 0.08. For (aii) I would use that to find the probability of none, exactly 1, and exactly 2. The probability of "more than 2" is 1 minus (the probability of 0 plus the probability of 1 plus the probability of 2).

For (bi) a "tree diagram" is a series of arrows showing the different possible outcomes. Here, from the "root" of the tree, you have two arrows ("branches"), that a baby is underweight and that a baby is not underweight. From each of those there are two more arrows, that the mother is a smoker and that the mother is not a smoker.

For (bii) the way I prefer to do it is this: Imagine 10000 babies (too avoid fractions). 16%, 1600, are underweight, 10000-1600= 8400 are not. Of the 1600 underweight babies, 58%, 928, have mothers who smoke. Of the 8400 who are not underweight, 27%, 2268, have mothers who smoke. That is a total of 2258+ 928= 3196 babies with mothers who smoke of which 928 are underweight (I am assuming one baby per mother).

 

[Old Student]

Thanks HallsoIvy's help

For a(i) , if I assume there are 100000 in city X, that means there are 80 people having respiratory disease! Right? Furthermore, can I think via this way?
Exactly 3 sick people can be selected, the probability is (80/100000) X (79/99999) x (78/99998) ~= 0.0008x0.0008x0.0008=0.000000000512

 

[HallsofIvy]

Thanks HallsoIvy's help

For a(i) , if I assume there are 100000 in city X, that means there are 80 people having respiratory disease! Right? Furthermore, can I think via this way?
Exactly 3 sick people can be selected, the probability is (80/100000) X (79/99999) x (78/99998) ~= 0.0008x0.0008x0.0008=0.000000000512

No, .08*100000 is NOT 80. Were you thinking that 0.08%= .0008? That is true but this problem says "the proportion of people having a respiratory disease is 0.08", NOT ".08 %".