The answer depends on the demography of the population from which the person is randomly selected.
The answer depends on the demography of the population from which the person is randomly selected.
The answer depends on the demography of the population from which the person is randomly selected.
The answer depends on the demography of the population from which the person is randomly selected.
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".
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.
27/35
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.
----------------------------------------------------------------------------------------------------------- ANSWER: Depends on the first supervisor's age and interests. -----------------------------------------------------------------------------------------------------------
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".
0.13 to 2 d.p.
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.
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 probability that a single person would like at least ONE flavour - is 9/10 * * * * * No. 350 liked only Vanilla 250 liked Vanialla and Chocolate 50 liked only Chocolate That makes 650 [the remaining 350 did not like either]. Therefore the probability that a randomly selected person likes at least one of the two tastes is 650/1000 or 65%
66%
There are 475 people total 300 are male so 175 are female (assuming they all identify as male or female) 135 males like fried food 270 people total do not like fried food 105 females do not like fried food 175 - 105 = 70 females do like fried food 270 - 105 = 165 males do not like fried food a.) Probability that the person is female is 175/475 = ~37% b.) Probability that the person is female and likes fried food is 70/475 = ~15% c.) Probability that the person is female or does not like fried food is (175 + 165) / 475 = 85%
27/35
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.
3/10 or 0.3
It is possible for the captain and the goalkeeper to be the same person. This changes the probability very significantly. There is nothing in the question to indicated that this is or is not the case.