In a software company 4/10 of people know C++, 2/4 of them know java, 2/8 of them know neither,find the total possibility to know C++?

• ans is 1/4....
  possibility of knowing c++= 4/10
  possibility of knowing java=2/4
  
  there are 4 possibilities,
  a person knows java but not c++=2/4*(1-(4/10))=3/10
  a person knows c++ but not java= 4/10*(1-(2/4))=1/5
  possibility of knowing neither =2/8
  therefore, probability of knowing any one language only= 3/10+1/5=1/2
  therefore ,possibility of knowing both= total probability- probability of knowing neither- probability of knowing anyone language only
  total probability is always 1
  so, required probability= 1 -(2/8) -(1/2) =1/4
  
• 7 years agoHelpfull: Yes(29) No(5)
• c++ known= 4/10
  java known=2/4
  neither=2/8
  both known=1-2/8=6/8
  Only c++= 6/8(both)-2/4=2/8=1/4
• 7 years agoHelpfull: Yes(21) No(0)
• Assume 100 persons in a company
  person knows c++=4/10=100*4/100=20 persons
  person knows java=2/4=100*2/4=50 persons
  neither=2/8=100*2/8=25 persons
  according to our assumption 100 person=20+50+25=95 person covered(5 remaining)
  so,total possibility=20 person(c++)+5(remaining)=25 persons
  which is 25/100=1/4
  
  
  
• 7 years agoHelpfull: Yes(12) No(19)
• c++ knows 2/5
  java knows 1/2
  knows nothing(jon snow :p) 1/4
  if total people = 20(LCM of 5,2,4)
  c++ knows = 8
  java knows =10
  knows nothing = 5
  therefore, (Java)U(C++) =(20-5)=15
  knows both the language(intersection of java and c++) = (10+8)-15=3
  therefore possibility to know c++ = 8+3=11
  ANS is 11/20
• 7 years agoHelpfull: Yes(5) No(5)
• it tells you in the question.... the opening statement - therefore 4/10
• 7 years agoHelpfull: Yes(2) No(5)
• 7/30
  n(c++unionjava)=n(c++)+n(java)-n(c++intersectionjava)
  n(c++unionjava)=4/10+2/4-6/8
• 7 years agoHelpfull: Yes(1) No(0)
• 1/2...........
• 5 years agoHelpfull: Yes(0) No(0)