In a room there are 7 persons.
The chance that 2 of them were born on same day of the week

• The answer is 1/7.
  The first person can have the birthday in any of the days of the week which gives the probability as 1( ie 7/7).
  The second person should have the birthday on the same day as the 1st ones, which decreases the sample spaces to 1day(The birthday of the 1st person), making the probability 1/7. hence total probability = 1*1/7=1/7.
  OR
  We can the birthday problem's solution, NO 2 PERSONS CAN HAVE THE SAME BIRTHDAY gives
  1(1-1/7) = 6/7.
  Now the required probability is the chance that 2 of them were born on same day of the week making it 1-6/7=1/7
• 11 years agoHelpfull: Yes(20) No(2)
• please correct it 1/7 can never be the answer...its 2160/(7^5)
  procedure is
  7C2 * 7/7 * 1/7 * (6/7)*(5/7)*(4/7)*(3/7)*(2/7)
  ^^selecting 2 among those 2nd person and the rest will be left
  persons who 2 one can born on exactly with 6,5,4,3,2 days to choose
  born on same born on any that day
  day day
• 11 years agoHelpfull: Yes(8) No(0)
• total 7 persons so let us take p1,p2,p3,p4,p5,p6,p7
  thep can form 7C2 pairs (p1p2,p1p3,p1p4,p1p5,p1p6,p1p7,p2p3,p2p4,p2p5,p2p6,p2p7.........)=21pairs
  so each pair can born in any of 7 days so total probability=21*7
  let us take from 7 pairs 1 pair of persons as born in same date so 7C1=7
  so (7)/(21*7)=1/21
  ans =1/21
  if wrong correct me pls
  
• 11 years agoHelpfull: Yes(1) No(0)
• 1-((7c1*6c1)/(7c1*7c1))=1-6/7
  =1/7 ans
• 11 years agoHelpfull: Yes(1) No(1)
• m4maths messed my description....saying it again...first select 2 persons from 7 who born on same day...then..one of those 2 will will have 7 options while the other will have exactly 1 option(which is chosen by 1)and rest wi be eft with 6,5,4,3,2 option to chose their birth day
• 11 years agoHelpfull: Yes(1) No(1)
• 2/7
  
  Probability of having the two person born on same day is 2 out of 7.
• 11 years agoHelpfull: Yes(0) No(3)