set A {1,3,10,17,19} and set B {9,12,15,18,21} and set C {4,7,10,13,16,19} probabilty of 'a num from A+ a num from B>a num from C

• Total of 5*5*6 cases,where are 17 false cases
  1+9!> 10,13,16,19
  1+12!> 13,16,19
  1+15!> 16,19
  1+18!> 19
  similarly 3+9!> 13,16,19
  3+12!> 16,19
  3+15!>19
  10+9!>19
  so probability=1-(17/150)=133/150
• 11 years agoHelpfull: Yes(30) No(2)
• out of a total of 5*5*6 cases,there are 19 false events
  1+9!> 10,13,16,19
  1+12!> 13,16,19
  1+15!> 16,19
  1+18!> 19
  similarly 3+9!> 13,16,19
  3+12!> 16,18,19
  3+15!>19
  10+9!>19
  so probability=1-(19/150)=131/150
• 11 years agoHelpfull: Yes(11) No(14)
• we can add num from a and num from b in 25 ways
  so total possible ways are 25
  by adding any num from a and any num from b will be greater than 4
  so ans is 25/25=1
• 11 years agoHelpfull: Yes(6) No(8)
• I think the probabiity is 1 according to the question.
  bcoz any num from A + any numb from B >> than atleast 4 and 7 of set C
• 11 years agoHelpfull: Yes(5) No(3)
• ans- 9/25
  {(1,9),(1,12),(1,15),(1,18)
  (3,9).....}
  there will be total 7 cases which satisfy and total 25 cases
  
• 11 years agoHelpfull: Yes(0) No(0)
• 1/6..
  because total no. of possible outcomes is 25*6
  and total no. of favourable outcomes is 25.
• 11 years agoHelpfull: Yes(0) No(0)