If N=3, there are 33 = 27 ways the three objects can be placed into the three buckets. For each bucket, see if you can verify that There is a probability of 8/27 that this bucket contains zero objects There is a probability of 12/27 that this bucket contains exactly one objects There is a probability of 6/27 that this bucket contains exactly two objects There is a probability of 1/27 that this bucket contains exactly three objects When each of N objects is randomly placed into one of N buckets, let P(N) be the probability that a bucket contains exactly 0, 1, or 2 objects. For example, we can use the calculations above to determine that P(3) = 8/27 + 12/27 + 6/27 = 26/27 = 96.3%. Determine P(100).
Contingency Table
A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.
Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
Suppose you have N objects and N buckets.
For each object, randomly place it into a bucket, with each of the N options being equally likely.
Through this random process, some buckets may end up empty, while other buckets may have multiple objects.
If N=3, there are 33 = 27 ways the three objects can be placed into the three buckets. For each bucket, see if you can verify that
- There is a
probability of 8/27 that this bucket contains zero objects - There is a probability of 12/27 that this bucket contains exactly one objects
- There is a probability of 6/27 that this bucket contains exactly two objects
- There is a probability of 1/27 that this bucket contains exactly three objects
When each of N objects is randomly placed into one of N buckets, let P(N) be the probability that a bucket contains exactly 0, 1, or 2 objects. For example, we can use the calculations above to determine that P(3) = 8/27 + 12/27 + 6/27 = 26/27 = 96.3%.
Determine P(100).
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