There are 58 persons sitting in a room, what is the probability that at least 2 of them will have the
birthday on the same day?
A) 0.50
B)0.76
C)0.99
D)0.83
Answers
Given :- There are 58 persons sitting in a room, what is the probability that at least 2 of them will have the
birthday on the same day ?
A)0.50
B)0.76
C)0.99
D)0.83
Answer :-
→ Total person = 58
→ Total days in a year = 365
so, The first person can be born on any day, so his probability will be = 365/365
similarly,
→ The next person is now limited to 364 probability is = 364/365.
→ The third person probability will be = 363/365 .
then,
→ last person probability will be = (365 - 58)/365 = 307/365 .
therefore ,
→ Probability of no person has birthday on same day will be = (365/365) * (364/365) * (363/365) ________ (307/365) = 365! / (307! * 365⁵⁸) ≈ 0.24 .
hence,
→ The probability that at least 2 of them will have the
birthday on the same day = 1 - probability of no person has birthday on same day .
→ Required probability = 1 - 0.24 = 0.76 (B) (Ans.)
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Given : There are 58 persons sitting in a room,
To Find : Probability that at least 2 of them will have the birthday on the same day
A)0.50
B)0.76
C)0.99
D)0.83
Solution:
Probability that at least 2 of them will have the birthday on the same day = 1 - Probability that no one one have the birthday on the same day
There are 366 days in a year in case of leap year as any one can be born on 29 feb also
so 58 person can have birthdays in (366)⁵⁸ ways
58 person having difference birthdays can be inn ³⁶⁶P₅₈
( as here who has birthday on which date matters hence Permutations is used )
Probability that no one one have the birthday on the same day = ³⁶⁶P₅₈ / (366)⁵⁸
= 0.00845
Probability that at least 2 of them will have the birthday on the same day =
1 - 0.00845
=0.99155
≈ 0.99
probability that at least 2 of them will have the birthday on the same day is 0.99
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