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Using Java, Your friend claimed that it is not too difficult to pass an all MCQ based test by guesswork. Being a budding

Posted: Fri Jul 01, 2022 5:52 am
by answerhappygod
Using Java, Your friend claimed that it is not too difficult topass an all MCQ based test by guesswork. Beinga budding programmer (with a background in probability andstatistics J), you decided to writea Java program to put this claim to test. You know that if yourepresent the outcome of the testusing a random variable X, then X follows a binomial distributionwith parameters, k n and p.Here k is the number of successes in n trials and p is theprobability of success in a single trial. Inthis case, n represents the number of questions on the test, krepresents the minimum number ofquestions that you need to pass the test and p is the probabilityof getting a single question right.For a 10-question test, n = 10, k = 6 and p = ΒΌ. (You need 6/10correct answers to pass the testand assuming four choices per question, the probability of gettinga correct answer by guessworkonly is ΒΌ). Test your program n = 10.Binomial Distribution:If a random variable X follows the binomial distribution withparameters n ∈ N and p ∈ [0,1], wewrite X ~ B(n, p). The probability of getting exactly k successesin n independent Bernoulli trialsis given by the probability mass function:𝑓(π‘˜,𝑛,𝑝) = Pr(π‘˜;𝑛,𝑝) = Pr(𝑋 = π‘˜) = .π‘›π‘˜/ 𝑃!(1 βˆ’ 𝑝)"#!for k = 0, 1, 2, ..., n, where:.π‘›π‘˜/ = 𝑛!π‘˜!(𝑛 βˆ’ π‘˜)!is the binomial coefficient, hence the name of the distribution.The formula can be understood asfollows: k successes occur with probability pk and n βˆ’ k failuresoccur with probability (1 βˆ’ p)n βˆ’ k.However, the k successes can occur anywhere among the n trials, andthere are 5"!6 differentways of distributing k successes in a sequence of n trials.Use the following method to compute factorial:
static int factorial(int n){int fact = n;for (int i = n-1; i > 1; i--) {fact *= i;}return fact;}Test your program for 10 questions only.probabilityToPass10Questions = probabilityToPass6Questions +probabilityToPass7Questions +probabilityToPass8Questions + probabilityToPass9Questions +probabilityToPass10QuestionsYou could write a loop for the above or compute each probabilityseparately.No sample I/O provided