Latest TCS Aptitude Question SOLUTION: 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.

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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.

Asked In TCS swastika (8 years ago)

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Read Solution (6) :

Garima Pathak  (8 years ago)

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

Is this solution Helpfull? Yes (10) | No (4)  

vibhor juneja  (8 years ago)

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

Is this solution Helpfull? Yes (4) | No (1)  

anchal sharma  (8 years ago)

can any one explain ths more better

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Richa  (8 years ago)

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)

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saumya singh  (8 years ago)

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.

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Nikhil   (8 years ago)

https://www.khanacademy.org/math/probability/probability-and-combinatorics-topic/probability_combinatorics/v/birthday-probability-problem


follow this link..

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