The standard answer for school work is 1/7. However, the reality may be slightly different since in parts of the world, particularly where induced hospital births are relatively common, it is claimed that births are brought forward so that medical staff can spend their weekend with their own families.
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1 out of 7 I think so!
The probability of at least 1 match is equivalent to 1 minus the probability of there being no matches. The first person's birthday can fall on any day without a match, so the probability of no matches in a group of 1 is 365/365 = 1. The second person's birthday must also fall on a free day, the probability of which is 364/365 The probability of the third person also falling on a free day is 363/365, which we must multiply by the probability of the second person's birthday being free as this must also happen. So for a group of 3 the probability of no clashes is (363*364)/(365*365). Continuing this way, the probability of no matches in a group of 41 is (365*364*363*...326*325)/36541 This can also be written 365!/(324!*36541) Which comes to 0.09685... Therefore the probability of at least one match is 1 - 0.09685 = 0.9032 So the probability of at least one match is roughly 90%
All I can say is that it isn't zero. :D.
Leaving aside leap years, the probability is 0.0137
The probability with 30 people is 0.7063 approx.