a ticket is drawn from two hundred tickets numbered from 1to 200 , find the probability that the number is divisible by 2 or 3 or 5.

• A)numbers divisible by 2: 200/2=100
  B)numbers divisible by 3: 200/3=66
  C)numbers divisible by 5: 200/5=40
  counting twice
  
  AB)numbers divisible by 6: 200/6=33
  AC)numbers divisible by 10: 200/10=20
  BC)numbers divisible by 15: 200/15=13
  counting 3 times
  ABC)numbers divisible by 30: 200/30=6
  Total of numbers = A + B + C - AB - AC - BC + ABC = 100 + 66 + 40 - 33 - 20 - 13 + 6 = 146
  Probability is =146/200 or 73/100
• 8 years agoHelpfull: Yes(5) No(0)
• P(a=number divisible by 2) = 100/200 = 0.5
  P(b=number divisible by 3) = 1/3 = 0.67
  P(c=number divisible by 5) = 1/5 = 0.2
  P(a and b=number divisible by 2 and 3) = 0.5*0.67
  P(b and c=number divisible by 5 and 3) = 0.2*0.67
  P(a and c=number divisible by 2 and 5) = 0.5*0.2
  P(a and b and c=number divisible by 2 and 3 and 5) = 0.5*0.67*0.2
  P(A or B or C) = 0.5+0.67+0.2-(0.5*0.67) -(0.67*0.2) -(0.2*0.5) + (0.5*0.67*0.2) = 0.868
• 6 years agoHelpfull: Yes(0) No(0)