Negative Binomial Distribution? NEED ADVICE AND HELP!!!!

Assume City A has three parties, Party B, Party C, and Party D, with n1, n2, n2 citizens in them, respectively. Assume, for people in B, they will always support the motion by the government, while, for people in C, they will always be against the government. However, those in D will support the policy by the government randomly with probability p and will make their decision independently for each policy. If a citizen is randomly chosen who showed the support to the government twice successively, find the probability he/she will support the government again.

For this question, I was thinking of using the negative binomial distribution to solve. However, I am confused if the number of citizens are of any use at all.
This was what I thought;

success: 3
number of trials: 3
probability of supporting gov: p

Negative Binomial distribution:

2C2*p^3*(1-p)^0 + 2C2*(1)^3*(1-p)^0

Is this correct? I have a feeling that this is terribly wrong. Please help. Need advice.


TS Contributor
find the probability he/she will support the government again.
My naive answer would be, he must be from party B or party C, so (n1*1,0 + n2*p)/(n1+n2) .

But maybe the information that s/he has already supported the government twice in a row
must be taken into account; the probability that s/he is from B surely increases, given this
information. Do you use Bayes' theorem in your training course?

Just my 2pence



Ambassador to the humans
I think Karabiner meant to say B or D. I don't agree with Karabiner's naive answer but do agree that Bayes' theorem should be applied here.


TS Contributor
If a citizen is randomly chosen who showed the support to the government twice successively

You are sampling randomly from share of population who has voted supportively two times....
What share of the population will vote supportively two time? All of type C so that must be n2/(n1+n2+n3) and what about type D? p^2*n3/(n1+n2+n3)

Call thes probabilities P1=n2/(n1+n2+n3) and P2=p^2*n3/(n1+n2+n3)

Since you know you are sampling from this specific share of population the probability of drawing type C person is P1/(P1+P2) and probability of drawing type D is P2/(P1+P2)

There are two ways a person who has supported twice can support again: 1) Being type C and deterministically supporting again 2) Being type D and stochastically supporting again with prob. p
I would guess the sum of these probabilities are what you are looking for...I have found my best guess but I will not rob you of the pleasure of finding the answer youreself. I implicitly used Bayes theorem...