Expert: Josh - 3/2/2011

Question
Hello- If there are 1000 black beans in a jar and 1 green bean, and 999 people pull beans from the jar, what is the probability of one person pulling the green bean? May I see the answer and the equation that leads to the answer, please?

Answer
Hi Tim,

To answer this problem, we can ask the opposite question, what is the probability that no one pulls out a green bean.

For the first person, the probability of picking a black bean (avoiding the green bean) is p(1)=1000/1001. After a black bean is removed, the probability of picking a black bean for the second person is p(2)=999/1000. Repeating this the probability of picking a black bean for the 3rd, 4th, 5th....person are p(3)=998/999, p(4)=997/998, p(5)=996/997....respectively.

Since each selection is independent, the probability that this sequence of events actually takes place is given by the product of the individual probabilities. That is, p(1)*p(2)*p(3)*...*p(999).

The interesting thing is that the majority of the terms collapse (cancel out because the same number appears in both the numerator and denominator). Thus,

(1000/1001)*(999/1000)*(998/999)*(997/998)*......*(3/4)*(2/3) = 2/1001.

[Note 1: observe the 1000 (top) cancels with the 1000 (bottom), same goes for 999 and so forth]

[Note 2: We need not go through 999 selections explicitly to work out the probability for the last term. Simply look at the numerator and study the pattern. For the 3rd term, the numerator is 998 and we started with 1000 in the 1st term. The difference is 2. By induction, the numerator for the Nth term must be 1000-(N-1), where N=999.]

Finally, how does this probability relate to the original question. Remember, we want to find the probability that any one ends up picking a green bean. This is the complement of what we obtained. We posed the question as "everyone avoiding the green bean" whereas what the question asked for is "anyone picking the green bean". So, the probability is 1-(2/1001) = 999/1001.

Hope this helps.