If you fall from any given height, whether or not you get hurt will depend on a large number of factors: what you fall on, how you fall, whether or not there was something that broke your fall on the way down and so forth.
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The probability that the birthdays of five persons chosen at random will fall in twelve different calender months is zero. You would need at least twelve persons to have a non zero probability.
2/7, as there 7 days in a week.
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%
You must notice that it is easier to find the probability that none of the dice are six. This is the complement of the answer. Pc= (55)/(65) = .4019 *the first die has 5 ways to not equal six, and so on for the remaining dice. [55] outcomes have no six. *the total number of outcomes gives each die 6 ways to fall. [65] outcomes overall. *Probability is the favorable outcome divided by the total number of outcomes. *Probability of an event is 1 minus the complement of the event. P = 1 - .4019 P = .5981 <<<<<<FINAL ANSWER
It is 1, since it falls on a Wednesday.