No, the answer is not just 2/25.
I will assume a 365-day year (and no, the answer is not 25/365 either).
I will also assume that the problem is asking for the probability that *at least* two have the same birthday.
First find the probability of all different birthdays, then subtract this from 1 in the final step.
The 1st person's birthday can be any date.
The 2nd person's birthday differs from the first one's birthday with probability 364/365.
The 3rd person's birthday differs from the first 2 birthdays with probability 363/365.
The 4th person's birthday differs from the first 3 birthdays with probability 362/365.
.
.
.
The 25th person's birthday differs from the first 24 birthdays with probability 341/365.
The probability of all 25 having different birthdays is
(365/365)(364/365)(363/365)(362/365)..… = 0.4313 .
The probability that *at least* two have the same birthday is 1 - 0.4313 = 0.5687 .
Lord bless you today!
I will assume a 365-day year (and no, the answer is not 25/365 either).
I will also assume that the problem is asking for the probability that *at least* two have the same birthday.
First find the probability of all different birthdays, then subtract this from 1 in the final step.
The 1st person's birthday can be any date.
The 2nd person's birthday differs from the first one's birthday with probability 364/365.
The 3rd person's birthday differs from the first 2 birthdays with probability 363/365.
The 4th person's birthday differs from the first 3 birthdays with probability 362/365.
.
.
.
The 25th person's birthday differs from the first 24 birthdays with probability 341/365.
The probability of all 25 having different birthdays is
(365/365)(364/365)(363/365)(362/365)..… = 0.4313 .
The probability that *at least* two have the same birthday is 1 - 0.4313 = 0.5687 .
Lord bless you today!
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Obviously 2/25 or 8%