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Activity         2


           Suppose, according to the latest police reports, 80% of all petty crimes are unresolved, and in your town,
           at least three of such petty crimes are committed. The three crimes are all independent of each other. From
           the given data, what is the probability that one of the three crimes will be resolved?
           Solution
           The first step in finding the binomial probability is to verify that the situation satisfies the four rules of
           binomial distribution:
           Number of fixed trials (n): 3 (Number of petty crimes)
           Number of mutually exclusive outcomes: 2 (solved and unsolved)
           The probability of success (p): 0.2 (20% of cases are solved)
           Independent trials: Yes
           Next:
           We find the probability that one of the crimes will be solved in the three independent trials. It is shown as
           follows:
           Trial 1 = Solved 1 , unsolved 2 , and unsolved 3 rd
                                        nd
                           st
                  = 0.2 × 0. 8 × 0.8
                  = 0.128
           Trial 2 = Unsolved 1 , solved 2 , and unsolved 3 rd
                              st
                                        nd
                  = 0.8 × 0.2 × 0.8
                  = 0.128
           Trial 3 = Unsolved 1 , unsolved 2 , and solved 3 rd
                              st
                                          nd
                  = 0.8 × 0.8 × 0.2
                  = 0.128
           Total (for the three trials):
                  = 0.128 + 0.128 + 0.128
                  = 0.384



        Poisson Distribution
        Poisson distribution is a probability distribution that is used to display the frequency of an event over a given time
        period. It is used to model count-based data like how many emails you receive in your mailbox in an hour or how
        many people enter a store in a day.
        The Poisson distribution is parametrised by the expected number of events λ (pronounced “lambda”) in a time
        or space window. The distribution is a function that takes the number of occurrences of the event as input (the
        integer called k in the next formula) and outputs the corresponding probability (the probability that there are k
        events occurring).
                      k –λ
                    λ e
          Poi(k;λ) =
                      k!
        The formula of Poi (k; λ) returns the probability of observing k events given the parameter λ corresponds to the
        expected number of occurrences in that time slot.

        2.2.2. Continuous Probability Distributions

        All outcomes greater than 0 (which would include numbers whose decimals continue indefinitely, such as pi =
        3.14159265) are used to create a continuous distribution. The foundations of probability theory and statistical
        analysis are the ideas of discrete and continuous probability distributions and the random variables they describe.


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