14. Consider the following questions: give the distribution name, carefully define the random variables you use and find the final answer. a) A professor has made 20 exams of which 8 are hard, 7 are reasonable, and 5 are easy. The exams are mixed up and the professor selects 4 of them at random to give to four sections of the course she is teaching. What is the probability that no sections receive a hard test

the probability that no sections receive a hard test is 0.1022

Step-by-step explanation:

Given that;

total number of exams = 20

hard exams = 8

reasonable = 7

easy exam = 5

so total exams that are bot hard = 7 + 5 = 12

we will make use of the probability mass function of hyper geometric distribution;

so the probability that no section of the received a hard test can be determined as follows;

P( x=0; N=20, n=4, k=8 ) = [ ⁸C₀ × ²⁰⁻⁸C₄₋₀ ] / ²⁰C₄

= [1 × ¹²C₄ ] / ²⁰C₄

= [1 × (12!/(4!×(12-4)!) ] / (20!/(4!×(20-4)!)

= [1 × 495 ] / 4845

= 495 / 4845

= 0.1022

Therefore the probability that no sections receive a hard test is 0.1022

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation