Expert: Josh - 3/22/2009

Question
what is the probabilty that at least 2 people in a random group of 3 will have the same sign of the Zodiac? Also calculate the theoretical probabilty.

Answer
Hi Jim,

The theoretical probability is the only thing we can calculate here, without conducting an actual experiment.

Task:
We want the probability that "at least" 2 people in a random group of 3 will have the same sign of the Zodiac. This is made up of two components.
(i) That 2 out of 3 have the same sign;
(ii) That 3 out of 3 have the same sign.

Key points:
- Associated with each person we select, there are two possible outcomes.
- To save typing this all over again, let us use "B" to denote the event that the person belongs to the zodiac sign we are interested in. This carries a probability of "p". Similarly, we use "N" to denote the opposite event that the person does NOT belong to the zodiac sign. This carries a probability of "q=1-p".
- We have p=1/12 and q=1-p=11/12 since there are twelve zodiac signs.

Using the binomial theorem:
From M selections, the probability of getting k people of type "B" and (M-k) people of type "N" is given by
   P(M,k) = C(M,k) * p^k * (1-p)^(M-k)
where the number of possible combination C(M,k)=M!/[k!(M-k)!],  factorial M!=M*(M-1)*(M-2)*....*2*1. By definition, 0! equals 1.

Illustration: (some detailed explanation if you are confused by the formula)
To shed light on this formula, consider requirement (i) above. First, let us deal with the counting aspect. To come up with two "B"s and one "N", one of three situations must occur.
We either get "B,B,N", "B,N,B" or "N,B,B". [In this notation, "B,B,N" describes a sequence of events. "B,B,N" means that the 1st person we pick "Belongs", the 2nd person we pick also "Belongs", but the 3rd person does "Not belong"]. The three situations all yield the same combination, but the order of selection is different in each. Observe that C(M,k)=3!/[2!*(3-1)!]=(3*2*1)/[(2*1)*(1)]=3 is exactly the number of combination we predict would arise, when we seek the number of ways of getting 2 "B" and 1 "N".
Second, we consider the probability that one of the three situations (either "B,B,N", "B,N,B" or "N,B,B") would arise. The permutation of order does not affect the calculation; all three situations have the same probability. Take "B,B,N" for example. The probability of getting a B from the first person is (p=1/12). Next, we select the second person. This, in itself, also carries a probability of p. Since the two people are selected independently, the joint probability of both events happening (that the 1st is B, AND the 2nd is B) is given by p*p. The probability of getting an "N" is q=1-p. Thus, to get first a "B", then a "B", followed by an "N", we multiply the individual probabilities together to get the JOINT probability P("B,B,N")= p * p * (1-p). This is identical to the "p^k (1-p)^(M-k)" part of the formula, where M=3 (three people we have to choose) and k=2 (number of B's we want). So, this gives some insight into the formula. You can do the same to verify the validity of the formula for requirement (ii), where M=3, k=3 and the only situation is "B,B,B".

Answer: For (i) P("B,B,N" or "B,N,B" or "N,B,B")=3*(1/12)^2*(11/12)  [plug M=3,k=2,p=1/12 into the binomial formula]. For (ii) P("B,B,B")=1*(1/12)^3. The final answer is the sum of the probabilities for (i) and (ii), viz., 3*(1/12)^2*(11/12)+(1/12)^3. I think it evaluates to  0.0197 (close to 2%)

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A note regarding "Probability Simulation"

The word "simulation" suggests that you are running a computer program to emulate a random process according to certain assumptions about the probability distribution. This is usually done for complex cases where a probability distribution might not have closed form, or the probability *space* is extremely large, that some form of statistical sampling is required to generate representative samples. I am sure this is not what is required in this instance, except, may be you could do some kind of survey, say on 1000 samples each consisting of 3 people, to work out the empirical probability. You produce a frequency histogram. This is essentially a counting exercise.