It is approx 0.001824
There is not enough information about the the distribution of the number of people known by each individual - nor the averages. It is therefore no possible to give an answer any more precise than "the probability will be infinitesimally small".
The probability that both will be hopelessly romantic is .0081 .009^2 = .0081
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.
people eat chees often
Since there are 73 Women, and 18 men, there are 91 people all together. The probability of the student being a man is 18 men out of the 91 total people. So, 18/91 or .1978.
If the events can be considered independent then the probability is (0.7)4 = 0.24 approx.
There is not enough information about the the distribution of the number of people known by each individual - nor the averages. It is therefore no possible to give an answer any more precise than "the probability will be infinitesimally small".
The probability that both will be hopelessly romantic is .0081 .009^2 = .0081
The answer would depend on the demographics of the population: a probability of 0.2 it too high unless the population is from a retirement area.
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.
people eat chees often
If I understand your question, yes, the proportion of people in a population ill with a certain disease at a given time is the same as the probablility that a randomly selected person in that population will have the disease at that time.
Since there are 73 Women, and 18 men, there are 91 people all together. The probability of the student being a man is 18 men out of the 91 total people. So, 18/91 or .1978.
27/35
The probability is approx 0.81
We will use Q and P to help solve this problem with Q representing the possibility that none of the randomly selected people are vaccinated and P representing the possibility that at least 1 randomly selected person is vaccinated. Because the sum of all probabilities must equal 1, your beginning equation will be P=1-Q. First you need to figure out how much of the population is NOT vaccinated so you would take 100%-54% to get 46%. With that 46%, you can conclude that any given person has the probability of 0.46 of not being vaccinated. To find the value for Q we will take (0.46)^5. Q=(0.46)^5=0.0206. To find P we go back to the original equation of P=1-Q. P=1-0.0206=0.9794. The probability that at least 1 person has been vaccinated is 0.9794.
1/365 = 0.00274