What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '5'?

• Total two digit numbers = 90
  The numbers multiple of 3 but not 5 = 12,18,21,24,27,33,36,39,42,48,51,54,57,63,66,69,72,78,81,84,87,93,96,99 = 24 numbers
  Hence probability = 24/90 = 4/15
• 12 years agoHelpfull: Yes(37) No(0)
• Since every third number starting from 10 will be divisible by 3, so total number of numbers divisible by 3 are 90/3 = 30
  
  Numbers which are divisible by 3 and 5 both are numbers which are multiple of 15.
  For the range 10 to 99, 15 is the first number divisible by 15 and 90 is the last number.
  So total number of numbers divisible by 15 are: (90-15)/15 + 1 = 5 + 1 = 6
  
  Number of numbers which are divisible by 3 are 30 and number of numbers which are divisible by 3 and 5 both are 6. So number of numbers divisible by 3 and not by 5 are: 30-6=24
  
  So total probability = 24/90 = 4/15
  
  
• 10 years agoHelpfull: Yes(3) No(0)
• total 2 digit numbers from 1 to 100 is 90
  the 2 digit numbers multiplied by 3 is 12,15,18,21,24.....99=30numbers
  so the probability is 30/90=1/3
• 13 years agoHelpfull: Yes(2) No(18)
• 10-30 number mul of 3 is 7 (15 il be div by 5 )so 7-1=6
  31-60 =10 (45..) 10-1 = 9
  61-90= 9
  93 96 99 = 3
  3+9+9+6 = 27
• 12 years agoHelpfull: Yes(1) No(5)
• total 2 digit nos=90
  multiples of 3 which are two digit=29
  multiples of 5 b/w these=6
  so 29-6=23
  hence total prob= 23/90!!!
• 12 years agoHelpfull: Yes(1) No(6)