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.
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%
1 out of 7 I think so!
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.
The probability of December 3 falling on a Friday is 1/7.
IF probability of rain is X percent then probability of no rain is 100- X percent. For example if prob of rain is 80% prob of no rain is 20%
1/7.
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%
1/7 chances
1 out of 7 I think so!
The probability of two people's birthday being the same is actually more likely than many would think. The key thing is to note that it doesn't matter what the first person's birthday is. All we need to work out is the probability that the second person has a birthday on any specific day. This probability is 1/365.25 The probability that they were born on June 10th is 1/365.25. The probability that they were born on February 2nd is 1/365.25 and the probability that they were born on the same day as you is 1/365.25
Rick Monday was born on November 20, 1945.
All I can say is that it isn't zero. :D.
monday
Leaving aside leap years, the probability is 0.0137
thirteen sixteenths.