1. ## Binomial Question?

In 1000 people in a company, the probability of having a disease is 0.005. One of the people is John and he knows he has the disease. What does John think is the probability that more than 1 person including himself has the disease?

I would denote J to be the number of people with the disease excluding john.

The answer is J~Bin(999, p) P(J>0) = 0.993

Can someone explain to me what is happening? How do you have the p?

2. Originally Posted by sankeel
In 1000 people in a company, the probability of having a disease is 0.005. One of the people is John and he knows he has the disease. What does John think is the probability that more than 1 person including himself has the disease?

I would denote J to be the number of people with the disease excluding john.

The answer is J~Bin(999, p) P(J>0) = 0.993

Can someone explain to me what is happening? How do you have the p?
Let X be the random variable "number of people ion company with disease".

X ~ Binomial(p = 0.005, n = 1000).

You have to calculate $\Pr(X \geq 2 | X \geq 1)$.

3. can somone give further explanation to mr. fantastic reply

4. Originally Posted by sankeel
can somone give further explanation to mr. fantastic reply
Which part of my reply don't you understand? Where is further explanation needed? Have you met the idea of conditional probability?

5. Hello, sankeel!

In 1000 people in a company, the probability of having a disease is 0.005.
One of the people is John and he knows he has the disease.
What is the probability that more than 1 person including John has the disease?

We have: . $\begin{Bmatrix} P(\text{disease}) &=& 0.005 \\ \\[-4mm] P(\text{"clean"}) &=& 0.995 \end{Bmatrix}$

What is the probability that John is the only one infected?

Then the other 999 people are "clean": . $P(\text{999 clean}) \:=\:(0.995)^{999}$

Therefore: . $P(\text{2 or more infected})$

. . . . . . . . $=\;P(\text{John is }not\text{ the only one})$

. . . . . . . . $=\;1 - (0.995)^{999}$