Probably probability! (2)

First, find the probability of all 50 persons having different birth dates!!

Thus, we got to consider only 365 days... the probability will be...

365/365 x 364/365 x 363/365......317/365 x 316/365=0.03

(the first person can be born on any date and any month of the year, the second one can choose only 364 dates out of 365 ....)

thus, the probability of 2 persons having same birth date will be:

1 - 0.03

= $\boxed{0.97}$

There are 365 days in a year, that's 365 possible birthdays(Leap not included). Therefore for two people the probability is: $1-\frac {364}{365}$ This is because there is one day they could share and 365 days total. Continuing decreasing for 50 people gives: $\frac { 364 }{ 365 } \times \frac { 363 }{ 365 } \times ...\times \frac { 317 }{ 365 } \times \frac { 316 }{ 365 } =\frac { 364! }{ { 365 }^{ 49 }\times 315! } =0.03$ However, we now need to subtract this answer from one to find the probability that one pair share a birthday. $1-0.03=\boxed{0.97}$ Now, there is a generalised formula for this: $\frac { 364! }{ { 365 }^{ n-1 }\times (365-n)! }$ Where n = the number of people choosing from! You can see why this is from the example above.

Credit should go to Jaiveer Shekhawat for filling in the gaps for some of my explanation. Thanks!