Latest TCS Aptitude Question SOLUTION: there are 7 people in a room,what is the probability that 2 of them hav same birthday

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(#M40029769) TCS Logical Reasoning question Keep an EYE

there are 7 people in a room,what is the probability that 2 of them hav same birthday

Asked In TCS (10 years ago)

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Rohit kumar prasad  (10 years ago)

number of days in a year be 365.
therefore probability of first person is 365/365.
probability of second person having diierent birthday is 364/365.
probability of third person having different birthday is 363/365.
similarly...
probability of seventh person having different birthday is 359/365.
total probability o seven people having different different birthdays is.
=(365/365)*(364/365)*(363/365)*(362/365)*(361/365)*(360/365)*(359/365).
therefore two people having same bithday=1-all having different birthdays.
=1-(365*364*363*362*361*360*359)/365^7.(ans).

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anirban kundu  (10 years ago)

365*365*364*363*362*361*360/(365)^7

Is this solution Helpfull? Yes (8) | No (3)  

Rohit kumar prasad  (10 years ago)

number of days in a year be 365.
therefore probability of first person is 365/365.
probability of second person having diierent birthday is 364/365.
=total probability is =(365*364)/(365*365).
therefore two people having same birthday=1-0.9972.
=0.0027(ans)

Is this solution Helpfull? Yes (3) | No (1)  

Sagnik Raha  (10 years ago)

this a problem of the famous birthday paradox!first of all we need to find out the probability of 2 persons having different birthday.First,we have a single person whose probability of birthday not matching any others is 365/365=1(as there is no previous analyzed person).For the second person we have probability of his birthday not matching with others as 364/365,for the 3rd person 363/365 and so on.Now here we have 7 persons in a room who can make 7C2 i.e 21 pairs.Hence,the probability of 2 persons whose birthday matches will be 1-(364/365)^21.The paradox is that when we have 23 persons in a room the probability of there birthdays matching is 50% whereas if we have 57 persons then the probability is 99.99%.This can be shown mathematically!

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rohit  (10 years ago)

7c2*(364^5/365^7)

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Sagnik Raha  (10 years ago)

the probability q(n) that someone in a room of n other people has the same birthday as a particular person is given by
q(n) = 1 - ((365-1)/(365))^n

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samarth shrivastava  (10 years ago)

as there are 7 persons and there are 7 days in a week. thus
probability of 1st person to have b'day on any one day =1/7
probability of 2nd person to have b'day on same day =1/7
probability of 3rd person to have b'day on some other day =6/7
likewise for others 5/7,4/7,3/7,2/7
thus ans is (1/7)*(1/7)*(2/7)*(3/7)*(4/7)*(5/7)*(6/7)=720/823543

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ABHRA  (10 years ago)

ans= (365-5)/365 = 72/73...

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