0
What is the probability of randomly selecting a sick person from a group of three sick people and seven healthy people?
Wiki User
∙ 11y ago
3/10 or 0.3
Wiki User
∙ 11y ago
Add your answer:
Earn +20 pts
You wish to use a long string of random digits to randomly assign one-half of a group of 12345 people to a treatment group You assign consecutive number labels to all the people starting with zero?
Quadruplets
What is the probability that a group of 6 people selected at random from 7 men and 7 women will have at least 3 women?
Using the hypergeometric distribution, the answer is 2114/3003 = 0.7040
How do you compute discrete variables?
You do not compute discrete variables. Some variables are discrete others are not. Simple as that. You do not compute people - you can compute their average height, or mass, or shoe size, etc. But that is computing those characteristics, you are not computing people. In the same way, you can compute the mean, variance, standard error, skewness, kurtosis of discrete variables, or the probability of outcomes, but none of that is computing the discrete variable.You do not compute discrete variables. Some variables are discrete others are not. Simple as that. You do not compute people - you can compute their average height, or mass, or shoe size, etc. But that is computing those characteristics, you are not computing people. In the same way, you can compute the mean, variance, standard error, skewness, kurtosis of discrete variables, or the probability of outcomes, but none of that is computing the discrete variable.You do not compute discrete variables. Some variables are discrete others are not. Simple as that. You do not compute people - you can compute their average height, or mass, or shoe size, etc. But that is computing those characteristics, you are not computing people. In the same way, you can compute the mean, variance, standard error, skewness, kurtosis of discrete variables, or the probability of outcomes, but none of that is computing the discrete variable.You do not compute discrete variables. Some variables are discrete others are not. Simple as that. You do not compute people - you can compute their average height, or mass, or shoe size, etc. But that is computing those characteristics, you are not computing people. In the same way, you can compute the mean, variance, standard error, skewness, kurtosis of discrete variables, or the probability of outcomes, but none of that is computing the discrete variable.
Is select an abstract noun?
Yes, the noun 'select' is a abstract noun, a word for chosen or preferred people or things; a word for a concept. The abstract noun form of the verb to 'select' is the gerund, selecting; a word for a process. The abstract noun form of the adjective 'select' is selectness; a word for a quality.
How many ways can 6 people be arranged in a line?
There are 6 ways of selecting the 1st person, 5 ways of choosing the 2nd person, 4 ways of picking the 4th person......and so on. The number of ways is 6 x 5 x 4 x 3 x 2 x 1 = 720.
A class of 14 boys and 11 girls. What is the probability of selecting a girl 1st and then a girl 2nd and then a boy third.?
Assuming the choices are made randomly and that the chosen people are not returned to the class, the probability is 77/690 = 0.1116 approx.
The probability of a person getting a cold is .7 What is the probability that 4 people if selected randomly will get a cold?
If the events can be considered independent then the probability is (0.7)4 = 0.24 approx.
Must randomly select 5 people out of 26 what is the probability that the 5 youngest are selected?
It is approx 0.001824
What is the formula if you have 20 people on a basketball team 45 people on a football team 10 are on both--how many ways can a person be selected from either team?
Let P(A) = 1/10; P(A) = probability of selecting one people on a basketball team P(B) = 1/35; P(B) = probability of selecting one people on a football team P(C) = 1/10 = probability of selecting one people who plays in both teams P(D) = probability of selecting from either team. P(D) = P(A) + P(B) - P(C) P(D) = 1/10 + 1/35 - 1/10 P(D) = 1/35 or 0.0286
About 9 percent of the population is hopelessly romantic If two people are randomly selected what is the probability both are hopelessly romantic?
The probability that both will be hopelessly romantic is .0081 .009^2 = .0081
If 4 percent of people commute by bicycle and a person is selected randomly what is the odds against selecting someone who commutes by bicycle?
24 - 1
What is the probability that one randomly-selected person in America will know another randomly-selected person in America?
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".
What is the probability that 4 randomly selected people all have different birthdays?
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.
What is the probability that in a randomly selected sample of 6 people exactly 4 of them are 65 approximately 0.2?
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.
What are the odds of having kids with the same birthday but a different year?
In probability theory, the birthday problem, or birthday paradox[1] pertains to the probability that in a set of randomly chosen people some pair of them will have the same birthday. In a group of 10 randomly chosen people, there is an 11.7% chance. In a group of at least 23 randomly chosen people, there is more than 50% probability that some pair of them will both have been born on the same day. For 57 or more people, the probability is more than 99%, and it reaches 100% when the number of people reaches 367 (there are a maximum of 366 possible birthdays). The mathematics behind this problem leads to a well-known cryptographic attack called the birthday attack. See Wikipedia for more: http://en.wikipedia.org/wiki/Birthday_paradox
A committee consisting of 6 people is to be selected from eight parents and four teachers Find the probability of selecting three parents and three teachers?
8/33
If On a Saturday evening 34 of the people in Chicago go out to dinner 18 see a movie 13 have a party and 35 stay home If seven people are randomly selected what is the probability that one eat?
people eat chees often
