Answer: 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