In a hotel, rooms are numbered from 101 to 550. A room is chosen at random. What is the probability that room number starts with 1, 2 or 3 and ends with 4, 5 or 6?

• Ans : 1/5
  XYZ x takes 3 nos (1,2,3) Y takes 10 values(0-9) Z takes 3 values (4,5,6)
  3*10*3=90
  prob = 90/450 = 1/5
• 13 years agoHelpfull: Yes(37) No(5)
• since the door number starts for 101 onwards,3rd digit is based on the first digit
  if the first digit is 1, then the total probability of 3rd digit lies in between 1-9 (0 should not be added since 100th room is not there)
  if the first room is 2 or 3, then 3rd digit have total probability of 10 (0-9).
  so if the 1st digit is 1, then probability is (1/5)*(10/10)*(3/9)
  if the 1st digit is 2 or 3, then probability is (2/5)*(10/10)*(3/10)
  the total probability for success is [(1/5)*(10/10)*(3/9)]+[(2/5)*(10/10)*(3/10)]=(1/15)+(3/25)= 14/75
  Ans is 14/75
• 13 years agoHelpfull: Yes(5) No(10)
• 90/450=1/5
  Ans:1/5
• 13 years agoHelpfull: Yes(4) No(3)
• for 1st place 1,2,3 can be taken in 3 ways and for last place 4,5,6 in 3 ways and for middle position any number so 10 ways so totally 90 numbers are there
  totally 550 numbers are there so probability is 90/550;
• 13 years agoHelpfull: Yes(3) No(14)