First we don't consider leap year
There is 1 month with 28 days: february
There are 4 months with 30 days: april, june, september, november
So the probability to select a month with 28 or 30 days is (1+4)/12 = 5/12
If the year is a leap year then the probability is 4/12
There are 12 months to choose from There are 7 months with 31 days in them. The probability of choosing a 31-day month is 7/12.
It is 1. Every month contains 28 days.
Absolute certainty
Let us assume that there are exactly 365 days in a year and that birthdays are uniformly randomly distributed across those days. First, what is the probability that 2 randomly selected people have different birthdays? The second person's birthday can be any day except the first person's, so the probability is 364/365. What is the probability that 3 people will all have different birthdays? We already know that there is a 364/365 chance that the first two will have different birthdays. The third person must have a birthday that is different from the first two: the probability of this happening is 363/365. We need to multiply the probabilities since the events are independent; the answer for 3 people is thus 364/365 × 363/365. You should now be able to solve it for 4 people.
The simplistic answer is that August will have five Mondays if the month starts on a Saturday, Sunday or Monday. Three days out of the seven so the probability is 3/7 = 0.4286. However, that assumes that any day of the week is equally likely and that is not the case. The impact of leap years and their 2800-year cycle is to increase the probability to 0.43
There are 12 months to choose from There are 7 months with 31 days in them. The probability of choosing a 31-day month is 7/12.
The probability is 7:12... There are five months with less than 31 days, so the probability of selecting a month with exactly 31 days is 7 out of 12.
there are 4 monthes that has 30 days each year there are 12 monthes each year. so the probabillity for picking a month that has 30 days in it is 4/12
It is 1. Every month contains 28 days.
It is 7/12.
The probability is approx 15/16.
Absolute certainty
you wont get your birthday every year
Just over 7 out of 12.
Let us assume that there are exactly 365 days in a year and that birthdays are uniformly randomly distributed across those days. First, what is the probability that 2 randomly selected people have different birthdays? The second person's birthday can be any day except the first person's, so the probability is 364/365. What is the probability that 3 people will all have different birthdays? We already know that there is a 364/365 chance that the first two will have different birthdays. The third person must have a birthday that is different from the first two: the probability of this happening is 363/365. We need to multiply the probabilities since the events are independent; the answer for 3 people is thus 364/365 × 363/365. You should now be able to solve it for 4 people.
It depends what you mean by randomly. If its not eaten anything for one or two days - but regains its appetite, I wouldn't worry. However - if it's been more than a couple of days - take it to a vet for a check up !
It depends on what month you select, since different months have different lengths.