Set A {1,3,10,17,19}, B{9,12,15,18,21} and set C{4,7,10,13,16,19} probability 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
• 9 years agoHelpfull: Yes(25) No(1)
• number of possibility [ set A + set B] > number of possibility of set C
  1 + 9 > {13,16,19} = 3!
  3 + 9 > {13,16,19}= 3!
  1 + 12 >{16,19} = 2!
  3 + 12> { 16,19}= 2!
  1 + 15 > { 19} = 1!
  3 + 15 > {19}= 1!
  total possibility from 6 combination = 2*3! + 2*2! + 1 + 1=18
  hence,
  probability = 6/18 = 1/3
• 9 years agoHelpfull: Yes(5) No(3)
• total no of cases - no of false cases=true cases
  total no of cases=150
  false cases=17
  1+9=10 which is not greater than 10,13,16,19 from set C. So this are the false case. Rest are true.
  Similarly calculate the other cases.
  likewise we will get 17 cases which are false. so,150-17=133 cases are true
  so the probability=133/150
  
• 9 years agoHelpfull: Yes(4) No(0)
• 12 is the answer
  
• 9 years agoHelpfull: Yes(1) No(8)
• total no of comaprision we have is 5c1*5c1*6c1=5*5*6=150
  total no of cases A+B>c is 12
  so probability=12/150
• 9 years agoHelpfull: Yes(1) No(0)
• total triplets we can get from A,B,C =5 elements from A *5 from B *6 from C =150
  table displays values of A+B
  AB| 9 12 15 18 21
  
  1| 10 13 16 19 22
  3| 12 15 18 21 24
  10| 19 22 25 28 31
  17| 26 29 32 35 38
  19| 28 31 34 37 40
  
  C={4,7,10,13,16,19}
  
  A+B>19 --->17 pairs * 6 match ups in C =102 triplets
  A+B>16 --->3 pairs * 5 match ups in C =15 triplets
  A+B>13 --->2 pairs * 4 match ups in C =8 triplets
  A+B>10 --->2 pairs * 3 match ups in C =6 triplets
  A+B>7 --->1 pairs * 2 match ups in C =2 triplets
  
  Total =133 triplets possible
  ==>p(A+B>C)=133/150
  
  
• 9 years agoHelpfull: Yes(1) No(0)
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• will anybody explain in a proper manner
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• had some confusion while solving this problem can any one explain it clearly
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• In set A -- there are 5 no: so we can select any no: from it in 5 ways
  Like wise,
  In set B -- there are 5 no: so we can select any no: from it in 5 ways
  In set C -- there are 6 no: so we can select any digit from it in 6 ways
  So, TOTAL WAYS in which 3 no:s can be selected such that one no: is selected from each set = 5 x 5 x 6
  = 150 ways
  now see the cases when (no: from A + no: from B) > no: from C
  i.e; let a be the selected element from A,
  let b be the selected element from B
  let c be the selected element from C
  then a + b should be greater than c
  It is simpler to calculate those cases where the condition fails
  i.e a+ b less than or equal to c
  such cases are :
  1+ 9
• 9 years agoHelpfull: Yes(0) No(0)
• such cases are :
  1+ 9
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• i'm sorry i have twice submitted the complete answer but its not being displayed.. sorry for the inconvenience
  
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