1. ## Probability

Could anyone help me with this by any chance... I am sure it is not that dificult for most people but I am really strugling ...

If 10% of the adult population are without debt at a particular time and a random sample of 20 people are chosen.

1. What is the probability distribution of the number of people WITHOUT DEBT?
2. What is the probability that NO MORE than 2 people are without debt?
3. What is the probability that at least 2 people are without debt?
4. How many people would be expected to be without debt?

2. Originally Posted by Grant Townend
Could anyone help me with this by any chance... I am sure it is not that dificult for most people but I am really strugling ...

If 10% of the adult population are without debt at a particular time and a random sample of 20 people are chosen.

1. What is the probability distribution of the number of people WITHOUT DEBT?
2. What is the probability that NO MORE than 2 people are without debt?
3. What is the probability that at least 2 people are without debt?
4. How many people would be expected to be without debt?
1. binomial b(n;20,0.1)

2. p(n<=2)=b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)

3. p(n>=2)=1-p(n<=1)=1-[b(0;20,0.1)+b(1;20,0.1)]

4. 0.1*20 =2

RonL

3. ## Probability

I do have another question and I hope you can help

How would I solve questions 2 and 3 without using excel?

4. Originally Posted by Grant Townend
I do have another question and I hope you can help

How would I solve questions 2 and 3 without using excel?
In your notes or text book you will find:

b(n;N,p) = (N!/(n!(N-n)!) p^n(1-p)^(N-n),

so:

2. p(n<=2)=b(0;20,0.1)+b(1;20,0.1)+b(2;20,0.1)

b(0;20,0.1) = 20!/(0!)(20!) (0.1)^0 (0.9)^20 = 0.9^20

b(1;20,0.1) = 20!/(1!)(19!) (0.1)^1 (0.9)^19 = 20*0.1* 0.9^19

b(2;20,0.1) = 20!/(2!)(18!) (0.1)^2 (0.9)^18 = (20*19/2)*0.1^2* 0.9^18