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

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

Vobo
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Takumi
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1 Answers1

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Hint: We find the probability that the person is in group B, given that she supported the government twice. If we know that conditional probability, then we know the probability she is in D (since she cannot be in C). And once we know these two probabilities, the answer to your question should be easy to get.

Let $V$ be the event she voted twice for the government, and let $B$ be the event she is in group B. We want $\Pr(B|V)$. By the definition of conditional probability, we have $$\Pr(B|V)=\frac{\Pr(B\cap V)}{\Pr(V)}.$$ We need the two probabilities on the right.

The event $V$ can happen in $2$ ways: She is in B, and she voted twice for the government or (ii) She is in D but voted twice for the government.

The probability of (i) is $\frac{n_1}{n_1+n_2+n_3}$. The probability of (ii) is $\frac{n_3}{n_1+n_2+n_3}p^2$.

A number of things have been left for you to do.

Instead of using the definition of conditional probability, you may prefer to use Bayes' Theorem.

Remark: We used $n_1,n_2,n_3$ for the populations. If you really meant $n_1,n_2,n_2$, just take $n_3=n_2$.

André Nicolas
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  • You are welcome. For a simpler problem of this type, we have a two-headed coin and a standard coin. We picked a coin at random and tossed it twice, and got heads. What is the probability we get a head on the third toss? – André Nicolas Feb 08 '14 at 07:44
  • P(B|V) = P(V|B)P(B) / P(V) = P(V|B)P(B) / [P(V|B)P(B) + P(V|D)P(D)]

    P(V|B) = n1/(n1+n2+n3) x 1 P(V|D) = n3/(n1+n2+n3) x p^2 P(B) = 1 P(D) = p

    After substitution I got: n1 / (n1 + n3p^3)

    Is there any mistake? Sorry I'm still very new to statistics

     – Takumi Feb 08 '14 at 07:46
  • You should get $\frac{n_1}{n_1+p^2n_3}$. Your calculation of $\Pr(B\cap V)$ was correct. The probability of $V$ is $\frac{n_1+p^2 n_3}{n_1+n_2+n_3}$. – André Nicolas Feb 08 '14 at 07:50
  • A comment: I have found that students who use Bayes' Theorem make more mistakes than students who use the definition of conditional probability. – André Nicolas Feb 08 '14 at 07:55
  • At first I found Bayes' Theorem very confusing especially with assigning the probabilities and distinguishing the probability such as between P(A|B) and P(A). When the probabilities are given as numbers I could handle it (still needs practice), but when it comes to alphabets such as this problem, it confuses me. – Takumi Feb 08 '14 at 08:03
  • By the way, when the smoke clears, you should find that the probability originally asked for is $\frac{n_1+p^3n_3}{n_1+p^2n_3}$. – André Nicolas Feb 08 '14 at 08:06