tickets number form 1 to 20 are mixed up and then a ticket is drawn at random what is the probability that ticket drawn has number which is a multiple of 3 or 5?

• the ways to draw 1 ticket from 20 =20c1=20
  total multiple of 3 or 5=(3,5,6,9,10,12,15,18,20)
  but 15 is multiple of both 3 and 5
  so it is not considered bcoz
  n(aub)=n(a)+n(b)-n(a intersection b)
  so probability=8/20=2/5
  
• 11 years agoHelpfull: Yes(2) No(0)
• no. of spaces=(1,2,3,4,5,...,19,20).
  
  event of getting a multiple of 3 or 5
  
  = (3,6,9,12,15,18,5,10,20)
  = 9/20. answer
• 11 years agoHelpfull: Yes(0) No(2)
• (6/20)+(4/20)-(1/20)=9/20
• 11 years agoHelpfull: Yes(0) No(0)
• numbers divisible by 3(3,6,9,12,15,18)=6/20
  numbers divisible by 5(5,10,15,20)=4/20
  15 is divisible by both 3 and 5 so 1/20
  6/20+4/20-1/20=9/20
• 11 years agoHelpfull: Yes(0) No(0)