- #1

- 15

- 0

## Homework Statement

In a given society, 15% of people have the sickness "S

_{a}" , from them 20% have the sickness "S

_{b}".

And from those that don't have the sickness "S

_{a}", 5% have the sickness "S

_{b}"

1-We randomly choose a person. and we define:

A:"the person having S

_{a}"

B:"the person having S

_{b}"

-Calculate: P(A) , P(B) and P(A∩B)

2-We take 10 persons from this society. We define X as the random variable that equals to the number of people having the sickness A and B at the same time.

-give the possible values of X, and the probability of them happening.

The first one is pretty easy, need some help with the second.

## Homework Equations

Results of the first question:

P(A∩B)=3%

P(A)=15%

P(B)=725/10000

## The Attempt at a Solution

X can take any value between 0 and 10.

I tried two things:

1- using the probability tree:

-P(X=0)=(1-P(A∩B))

^{10}≈0.73

-P(X=1)=10*P(A∩B)*(1-P(A∩B))

^{9}=0.22

But then it gets a bit too complicated to know how many times it's repeated.

2- We suppose that this society consists of 1000 people, so 30 of the will have both S

_{a}and S

_{b}:

-P(X=0)=C

^{10}

_{970}/C

^{10}

_{1000}=0.73

-P(X=1)=(C

^{1}

_{30}*C

^{9}

_{970})/C

^{10}

_{1000}=0.22

.

.

.

.

I think both are correct, but the first is way too complicated for the bigger ones.

And is there another method to avoid the supposition about the number of people in the society.