find probablity of a*b+c is even . numbers are from (1,2,3,4,5). digits can be repeated.

• given set is 1,2,3,4,5
  our aim is to find probability so that a*b+c should be even
  a*b+c gives even number when two odd numbers is added
  i.e, odd no+odd no gives even no.
  and even no + even no gives even no
  so lets consider case 1: odd+odd=even (a*b) is odd and c is odd
  now a*b gives a odd no when a and b are odd no.
  probability of odd number from given set=3 by 5
  so case 1 gives the prob as (3 by 5)*(3 by 5)*(3 by 5)=(27 by 125)........1
  
  case 2: even+even=even
  prob of even no from the set=2 by 5
  a*b is even and c is even (three cases are possible)
  a(odd) b(even) c(even)= (3 by 5)*(2 by 5)*(2 by 5)and
  a(even) b(odd) c(even)= (2 by 5)*(3 by 5)*(2 by 5)and
  a(even) b(even) c(even) = (2 by 5)*(2 by 5)*(2 by 5).
  so totally (27+12+12+8)by 125......ans
• 12 years agoHelpfull: Yes(35) No(1)
• for make a even number there will be the case
  1) a even b even c even ie prob = (2/5)*(2/5)*(2/5)=8/125
  2) a even b odd c even ie prob = (2/5)*(3/5)*(2/5)=12/125
  3) a odd b even c even ie prob = (3/5)*(2/5)*(2/5)=12/125
  4) a odd b odd c odd ie prob = (3/5)*(3/5)*(3/5)=27/125
  so tltal probability =(8+12+12+27)/125 ans
• 12 years agoHelpfull: Yes(19) No(1)
• its 59/125.... correct
• 12 years agoHelpfull: Yes(5) No(3)
• ans should be 99/125
  total cases =125
  cases for
  1- a (even ),b+c(even)
  a=2
  b+c=(3c1*3c1)+(2c1*2c1)=13
  total cases=2*13 =26
  
  2- a (odd ),b+c(even)
  a=3
  b+c=(3c1*3c1)+(2c1*2c1)=13
  total cases=3*13 =39
  
  3-a (even ),b+c(odd)
  a=2
  b+c=(3c1*2c1)+(3c1*2c1)=12
  total cases=2*12 =24
  
  now, p= (24+39+26)/125 = 99/125
• 12 years agoHelpfull: Yes(0) No(12)
• probability of even number=2/5
  prob of odd number=3/5
  total prob=(2/5)*(3/5)*(2/5)+(3/5)*(2/5)*(2/5)+(3/5)*(3/5)*(3/5)=51/125
• 12 years agoHelpfull: Yes(0) No(1)