# Thread: (Simple) Probability Question

1. ## (Simple) Probability Question

One of the three people in a group comes home with measles. Each of the other two people has a probability θ of catching measles from one who is infected. If neither or both get the measles, the epidemic ends.
However, if only one of them gets the disease, the remaining person has another opportunity θ, of being infected.

Let X be the total number of people who are infected before the epidemic
ends.

Show that

P(X=1) = (1 - θ)^2
P(X=2) = 2θ.(1 - θ)^2
P(X=3) = θ^2.(3-2θ)

Can anyone help me to start off?

2. Originally Posted by da`
One of the three people in a group comes home with measles. Each of the other two people has a probability θ of catching measles from one who is infected. If neither or both get the measles, the epidemic ends.

However, if only one of them gets the disease, the remaining person has another opportunity θ, of being infected.

Let X be the total number of people who are infected before the epidemic
ends.

Show that

P(X=1) = (1 - θ)^2
P(X=2) = 2θ.(1 - θ)^2
P(X=3) = θ^2.(3-2θ)

Can anyone help me to start off?
$\displaystyle \Pr(X = 1) = (1 - \theta)(1 - \theta)$ should be obvious since you require neither of the other two catching it from the one who does.

X = 2 means that you need one of the other two to catch it and then the third person NOT to catch it:

$\displaystyle \Pr(X = 2) = \Pr(\text{One of the other two catching it}) \cdot \Pr(\text{Third person doesn't catch it})$ $\displaystyle = 2 \theta (1 - \theta) \cdot (1 - \theta)$.

Pr(X = 3) = 1 - Pr(X = 2) - Pr(X = 1) - Pr(X = 0) = 1 - Pr(X = 2) - Pr(X = 1) since obviously Pr(X = 0) = 0.