in a gathering of 15 person none of them was born in a leap year. find the probability that at least two have birthday on same date.

• p(atleast two have same bdays) is very large to calculate
  so we will find out p(diff. birthdays)
  then p(same bdays)=1-p(diff. bdays)
  there are 15 peoples therefore
  p(different bdays)=365*364*363*.......*351/365^15
  p(diff.bdays)=365!/(365-15)!*365^15
  then
  p(same bdays)=1-365!/350!*365^15
  p(same bdays)=(350!*365^15-365!)/365^15 answer
  
• 8 years agoHelpfull: Yes(10) No(4)
• p(atleast 2 on same date)=1-p(none on same dates)
  p=(favourable/total)
  favourable=15 (15 people born on 15 diff dates)
  total=365 days
  hence, [1-(15/365)]=0.958
• 8 years agoHelpfull: Yes(4) No(1)
• can any one explain ths more better
  
  
• 8 years agoHelpfull: Yes(2) No(1)
• Let A be event at least 2 have birthday on same day so A' be event none have birthday on same day.Sample space will be 365*365*365 upto 15 times. As among 365 day 1 person b'dae can fall on any 1 day same with 2,3, nd all 15 person... So now when none of them have b'dae on same day we have 1 person b'dae among one day from those of 365,next 2 persons can b on remaining 364days, 3 person 364 continuing this for 15 person last value is 351.
  Thus we have A' = 365*364*363*362----------*351/365 ^15
  Probability for atleast 1- A'
  Simply this we have 365! /(350! *365^15)
  This is because once we solve 365!/350! We will have 365*364*363----------*351*350!/350! Where 350! Will b cancelled just to standardise we write it 1-365!/350!*(365^15)
• 8 years agoHelpfull: Yes(2) No(0)
• There are 15 peoples therefore,
  Probability of different birthday = (365 *364*363* ....... * 351)/(36515)
  
  Probability of different birthday = 365!/(365-15)! *36515)
  Then,
  The probability of same date = 1 - probability of different dates
  The probability of same date =1 - [365!/(365-15)!] *36515)
  The probability of same date,
  = (350!*36515 - 365!)/36515.
• 8 years agoHelpfull: Yes(1) No(0)
• https://www.khanacademy.org/math/probability/probability-and-combinatorics-topic/probability_combinatorics/v/birthday-probability-problem
  
  
  follow this link..
• 8 years agoHelpfull: Yes(0) No(0)