5. Find the probability that a 3 digit number formed by using the digits 1, 3, 6, 9 without repetition, is divisible by 4.
a. 1/2
b. 1/3
c. 1/4
d. 1/5

• if the number should be divisible by 4 then last two digits must be multiples of 4
  so the ways of locating unit digit is 3 ways out of (3,6,9)
  tenth digit is 2 ways
  hundredth digit is 1 way since repetition is not allowed
  total ways = 1*2*3 =6
  Total sample space= 4!
  probability = 6 / 24 = 1 / 4
• 9 years agoHelpfull: Yes(9) No(0)
• ans is (c)1/4.
• 9 years agoHelpfull: Yes(1) No(0)
• ans=1/2
  ( divisible by 4 implies the last two ns should be divisible by 4 i,e only 16 so only 1 way
  2c1*1/4c3=2/4=1/2)
• 9 years agoHelpfull: Yes(0) No(4)
• ANS =1/2
  divisible by 4 implies the last two nums should be divisible by 4 . that are 16 and 36 so two ways.
  hundredth palce disgit can be chosen in total 4 ways,
  so p=2/4=1/2
• 9 years agoHelpfull: Yes(0) No(2)
• divisible by 4 implies the last two nums should be divisible by 4 . that are 16 and 36 so two ways.
  there is no repetition means remaining 2 numbers ,2way
  total 2+2=4
  ans:4
• 9 years agoHelpfull: Yes(0) No(1)