chosen a,b,c randomly and with replacement from the set {2,3,4,5,6}the probablity that a*b+c is even ?

• ANSWER=18/25
  
  total number=5 numbers are {2,3,4,5,6}
  probability to pick a number p=1/5
  even numbers=3 {2,4,6} and Odd numbers=2 {3,5}
  probability of even number P.E=3/5
  probability of even number P.O=2/5
  given A*B+C is even
  case 1:
  if A is even then B can be either even or odd and C must be even to give even number
  P.E*p+P.O=
  3/5*1/5+3/5=18/25
  therefore probability=18/25
  case 2:
  if A and B are Odd then C must be odd to give an even number
  this case fails because there are only 2 odd numbers
  probability=0
  
  the probablity that a*b+c is even is p(case 1)+ p(case 2)=18/25+0= 18/25
  probability =18/25
  
  
  
  
  
• 10 years agoHelpfull: Yes(31) No(4)
• There are 4 cases possible
  P(e)=3/5 and P(o)=2/5
  
  Case1-All a,b,c are even
  so P1= (3/5)^3
  Case2- a,c = even and b=odd
  so P2=(3/5)(2/5)(3/5)
  Case3- b,c=even and a=odd
  so P3=(2/5)(3/5)(3/5)
  Case4- all odd
  so P4=(2/5)^3
  
  Adding all to get the total probability =71/125
  
• 10 years agoHelpfull: Yes(8) No(6)
• See number to be even...>
  {even*even+even}
  {even*odd+even}
  {odd*even+even}
  
  here, P(e)=3/5,P(o)=2/5
  so total probablity is.....>
  (3/5*3/5*3/5)+(3/5*2/5*3/5)+(2/5*3/5*3/5) = 63/125............hope it's correct.
• 10 years agoHelpfull: Yes(7) No(3)
• Probability of chosing a,b,c randomly and with replacement = 1/5
  Thus, a*b+c = 1/5*1/5+1/5 = 6/25
• 10 years agoHelpfull: Yes(1) No(7)
• (a,b,c)= (2,4,6)
  2*4+6=14
  2*6+4=16
  4*6+2=26 all are even
• 10 years agoHelpfull: Yes(0) No(3)
• Total ways of selecting 3 no.=5c3=10
  The sum willbe odd in 2 cases
  1_when all 3 nos r even whose probablity is 1
  2-when2 no. r odd nd 1 is even since only 2 no. r odd in this to select 1 even from 3 even no.s
  i.e. 3c1=3
  fAvourable cases =3
  thus p=4/10 ans=2/5.
• 10 years agoHelpfull: Yes(0) No(3)
• no of event=3C2*2C1 + 3C3
  no of sample space=5C3
  ans=7/10
  
  
• 10 years agoHelpfull: Yes(0) No(1)
• Answer=7/27
  there are 3 cases
  p1=all are even p2=a and c are even p3=b and c are even
  case 1:all are even
  p1=(1/3)*(1/3)*(1/3)=(1/27)
  p2=(1/3)*(1/3)=(1/9)
  p3=(1/3)*(1/3)=(1/9)
  p=p1+p2+p3=7/27
  
• 10 years agoHelpfull: Yes(0) No(1)
• 1-(6/pow(5,3))
  3 and 5 are odd. So, odd*odd+even=odd which is (a,b) either {(3,5),(5,3)}
• 10 years agoHelpfull: Yes(0) No(1)
• ((1/2)*(1/2))+1/3=7/12
• 10 years agoHelpfull: Yes(0) No(2)
• 71/125 is the correct answer
  As there are 4 cases:
  1.all odd ie (2/5)^ 3
  2.all even ie (3/5)^3
  3.a= odd other two even ie (3/5)^2* 2/5
  4.b= odd other two even ie (2/5)^2* 3/5
  Adding all we get 71/125
  
• 8 years agoHelpfull: Yes(0) No(0)