Discrete Probability Distributions

  • For a binomial distribution to be appropriate:

    • the number of trials are fixed and random samples

    • on each trial, the outcomes can be either success or failure

    • outcome of each trial is independent of the outcome of any other trial

    • the probability of success is the same on each trial

  • If given probability and asked to find n, just use the normal idea of probability to the power of n even if says use binomial distribution

  • Expectation for binomial distribution = np

  • variance for binomial = npq, q =1-p

  • For a Poisson distribution, the expectation(mean) and variance are the same, so if both binomial and poisson can be used to calculate probability this means mean = variance

  • P(X=r) = e-mean 2.5r / r!, very important if given probability and asked to find mean

  • Conditions for Poisson:

    • The variable is the frequency of events that occur in a fixed interval of time

    • events occur randomly(do not occur at regular or oredictable intervals)

    • Events occur independently/singly of one another - whether or not one even occurs does not affect the probability of whether another event occurs

    • events have to occur at a constant average rate

  • For the poisson graph, the distribution is positively skewed if the value of lamda i.e. mean, is very small; but as mean increases the distribution becomes progressively more symmetrical

  • When modelling data with Poisson distribution, the closeness of the mean and variance is one indication that the model fits the data well

  • Summing Poisson distributions by assuming both distributions are independent random variables with two separate means, then the sum Poisson distribution has the mean of the total mean

  • Linking binomial and poisson

    • the mean for poisson can be found using mean = np in binomial distribution, the results obtained from both are usually very similar

    • the smaller the value of p and the larger the value of n, the better the two distributions agree

  • Uniform distribution

    • in general, the distribution is defined as P(X=r) = 1/n, for r = 1,2,…,n, P(X=r)=0 otherwise

    • But lowest value may be k rather than 1, then the distribution over the values {k, k+1,…n+k-1}

    • expectation = (n+1)/2

    • variance = 1/12 *(n²-1)

    • question 6 in the paper - make always need to find the number of numbers and remember that probability is uniform

  • Geometric distribution

    • all about the probability of success and failure

    • for geometric, however many unsuccessful attempts makes no difference to how many more attempts she will need

    • conditions:

      • you are finding the number of trials it takes for the first success to occur

      • on each trail, the outcomes can be classified as either success or failure

      • the outcome of each trail is independent of the outcome of any other trail

      • the probability of success is the same on each trail

    • X~Geo(p) over the values {1,2,3…}

    • Expectation(mean) = 1/p, variance = (1-p)/P²