One digit is selected from first 30 positive integers , what is the probability that it is divisible by 3 or 5?

• total no. of events = 30
  numbers divisible by 3 are 3,6,9,12,15,18,21,24,27,30 a total no. of 10 numbers
  numbers divisible by 5 are 5,10,15,20,25,30 a total no. of 6 numbers
  no. of numbers divisible by 3 or 5 are [ (10+6)-2 ] = 14 {since 15,30 comes twice}
  HENCE probability is = 14/30 = 0.466667
  
  
• 12 years agoHelpfull: Yes(15) No(1)
• The probability that is divisible by 3 or 5 is 14/30 = 0.4666
  Total events=30
  feasible events for a number divide by 3=10(3,6,9,12,15,,18,21,24,27,30)
  feasible events for a number divide by 5=6(5,10,15,20,25,30)
  feasible events for a number divide by 3 and 5=2(15,30)
  
  
  The probability of number is divide by 3 =14/30
  The probability of number is divide by 5 =16/30
  The probability of number is divide by 3 and 5 =2/30
  so
  ((10/30)+(6/30)-(2/30))=14/30 =>0.4666
  
  
• 12 years agoHelpfull: Yes(5) No(1)
• 14.............p(a or b)=p(a)+p(b)-p(a and b)
  p(3)=10
  p(5)=6
  p(3 and 5)=2
  therfore 10+6-2=14..............
• 12 years agoHelpfull: Yes(4) No(0)
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• 12 years agoHelpfull: Yes(1) No(1)
• no divisible by 3= 30/3=10
  no divisible by 5=30/5=6
  no divisible by 3 and 5 = 30/15=2
  total no divisible by 3 or 5 =10+6-2=13
  total no=30
  probablity=14/30 = 0.467
• 8 years agoHelpfull: Yes(0) No(0)