Answer to Question #199328 in Statistics and Probability for Bahawal tahir

Question

The class registrations of 120 students are analyzed. It is found that: 30 of the students do not take any of Applied Mechanics, Chemistry, or Computers. 15 of them take only Applied Mechanics. 25 of them take Chemistry and Computers but not Applied Mechanics. 20 of them take Applied Mechanics and Computers but not Chemistry. 10 of them take all three of Applied Mechanics, Chemistry, and Computers. A total of 45 of them take Chemistry. 5 of them take only Chemistry. 

a) How many of the students take Applied Mechanics and Chemistry but not Computers? 

b) How many of the students take only Computers? 

c) What is the total number of students taking Computers? 

d) If a student is chosen at random from those who take neither Chemistry nor Computers, what is the probability that he or she does not take Applied Mechanics either? 

e) If one of the students who take at least two of the three courses is chosen at random, what is the probability that he or she takes all three courses?

Answer 1

$Total \space students \space =n \space = \space \space 120 \space \\ \space Students \space who \space not \space applied \space =30 \space \\ \space Students \space who \space applied \space = \space 90 \space \\ \space Only \space mechanics \space = \space 15 \space \\ \space Chemistry \space and \space computers \space but \space not \space Applied \space Mechanics. \space = \space \space \space 25 \space \\ \space Mechanics \space and \space computers \space but \space not \space Chemistry.= \space \space \space 20 \space \\ all \space three \space of \space Applied \space Mechanics, \space Chemistry, \space and \space Computers.= \space \space \space \space 10 \space \\ \space Chemistry \space = \space 45 \space \\ \space Only \space chemistry \space = \space \space 5 \space \\\ hence \space \\ (i)Mechanics \space and \space chemistry \space but \space not \space computers \space =45-(5+10+25)=5 \space ...eq(1)\\ (ii)only \space computers \space =90-(45+15+20)=10... \space eq(2)\\$

$\space a) \space How \space many \space of \space the \space students \space take \space Applied \space \\ \space \space Mechanics \space and \space Chemistry \space but \space not \space Computers? \space \\ \space Solution \space \\ \space with \space help \space of \space equation \space (1)\\ the \space students \space take \space Applied \space \\ \space \space Mechanics \space and \space Chemistry \space but \space not \space Computers \space =5\\ -----------------------\\ b) \space How \space many \space of \space the \space students \space take \space only \space Computers? \space \\ solution\\ with \space help \space of \space eq(2)\\ the \space students \space take \space only \space Computers=10\\ -----------------------\\ c) \space What \space is \space the \space total \space number \space of \space students \space taking \space Computers? \space solution\\ with \space help \space of \space diagram\\ the \space total \space number \space of \space students \space taking \space Computers=20+10+25+10=65\\ -----------------------\\ d) \space If \space a \space student \space is \space chosen \space at \space random \space from \space those \space \\ who \space take \space neither \space Chemistry \space nor \space Computers, \space \\ what \space is \space the \space probability \space that \space he \space or \space \\ she \space does \space not \space take \space Applied \space Mechanics \space either? \space \\ solution\\ total \space student \space who \space belong \space to \space chemistry \space or \space computer \space =45+20+10=75\\ total \space students=120\\ students \space who \space take \space neither \space chemistry \space nor \space computers=120-75=45\\ students \space who \space not \space take \space applied \space mechanics \space either=30\\ required \space probability=\frac{30}{45}=\frac{2}{3}\\ -----------------------\\ e) \space If \space one \space of \space the \space students \space who \space take \space at \space \\ \space least \space two \space of \space the \space three \space courses \space is \space chosen \space at \space random, \space \\ what \space is \space the \space probability \space that \space he \space or \space she \space takes \space all \space three \space courses?\\ solution\\ students \space who \space take \space at \space least \space two \space of \space three \space courses \space is \space chosen=20+25+5+10=60\\ students \space who \space takes \space all \space three \space courses= \space 10\\ required \space probability=\frac{10}{60}=\frac{1}{6}\\$