Question

A student completely guesses on every question on a 3-problem multiple choice quiz in which each question has five possible answers (p(correct on any one problem) = 0.20)). Let X be the number of problems that the student gets correct out of 3. Blank 1: For this to be binomial, BINS must be satisfied. Are there two outcomes (yes or no)? Are trials independent (yes or no)? Is this out of a fixed number of trials (identify n with a number)? Is the probability of success constant on each trial (identify p as a decimal to two places)? To report your answer for this discussion, give a list of 4 items matching what is described in the () of each question. For example: yes, yes,10,0.50 Blank 2: Determine the probability distribution of X, the number of problems correct out of 3 total. Note that X can take on values from O to 3 so you are finding P(O), P(1), P(2) and P(3). Report your answer as P(O),P(1),P(2),P(3) with no extra spaces. You may either report them as fractions, not simplified from what comes naturally in the problem, or as decimals to the needed places to keep them as exact answers. Blank # 1 Blank # 2

Answer 1

Facts: 12! 1) Binomial distribution : P(X = x) =- *p** (1-p)"-* Where n = total number of trials X = number of success's we chosen out of n p = probability of success. X = number of questions the student gets correct. Given that The number of questions n= 3. Blank 1) Yes, there are two outcomes either success or failure. Question gets correct comes under success and question gets wrong comes under failure.

Yes, trials are independent. One question outcome does not effect on remaining questions outcome. Yes, there are fixed number of trials. The number of Trials = The number of questions = n = 3. Yes, the probability of success constant on each trial. Each question has 5 options and only one option is correct. The probability that a randomly selected question gets correct

Reporting the answer: yes, yes,3,0.20. Blank 2) X satisfied all the porperties of binomial random variable. So X is a binomial random variable with n = 3 and p = 0.20 n ! Probability distribution of X:P(x) = x! (n-x). *p* * (1-pn-*

3! P(0) = 1.630), *(0.2)° * (1 – 0.2)3-0 = 1*1*(0.8)3 = (0.8)3 = 0.512 P(1) = .3-0), * (0.2)2 + (1 – 0.2)3–1 = ** (0.2)2 + (0.8) = 3 * (0.2) * (0.8) = 0.384 P(2) = 21.03.2), * (0.2)2 * (1 – 0.2)3-2 = 21.8)* (0.2)2 + (0.8)1 = 3 * (0.2)2 * (0.8)1 = 0.096

P(3) = 2.3-3); * (0.2)2 + (1 – 0.2) 3-3 3! + (0.2)3 + (0.8)º 3!-(0) = (0.2) = 0.008 Reporting the answer: 0.512,0.384,0.096,0.008 Blank #1: yes, yes,3,0.20 Blank #2: 0.512,0.384,0.096,0.008

Thank you.