Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?
Accepted Solution
A:
Answer:(b) 1/792Step-by-step explanation:The complete question is;Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.(a) How many different groups of 5 problems can be chosen from the 12problems?(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?In (a)Apply the formula[tex]\frac{n!}{(n-r)!(r!)}[/tex]where n=12 and r=5substitute values[tex]=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792[/tex]In (b)If Jerry did only 5 problems of one assignment then the probability will be[tex]\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}[/tex]