There is no information on the population from which the person is selected. But if, in a total population of N people, T drive trucks, E exercise four or more times per week, and TE are truck drivers who exercise four or more times per week, then the probability is
T/N + E/N - TE/N.
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 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.
----------------------------------------------------------------------------------------------------------- ANSWER: Depends on the first supervisor's age and interests. -----------------------------------------------------------------------------------------------------------
There being 365 days in a year and 50 being less than 365 therefore 2 even far less than that the chances are virtually 0. ______________________ Assume that all 366 days (including leap day) are equally likely to be a person's birthday. The probability that none of them share a birthday is 1*P(second person selected doesn't share a birthday with first person selected)*P(third person selected doesn't share a birthday with first or second person selected)*...*P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected). P(second person selected doesn't share a birthday with first person selected) = 365/366 P(third person selected doesn't share a birthday with first or second person selected)=364/366 . . . P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected)=317/366 P(none share a birthday)=(365/366)*(364/366)*...*(317/366), which is approximately .0299. P(at least two share a birthday) = 1-(365/366)*(364/366)*...*(317/366)=1-.0299=.9701 = 97.01%.
The probability that there are 1,2,3 or 4 men is 1-(the probability that no men are selected). First we select the first person. The probability that this person is a woman is 5/10=1/2. For second person it is 4/9, then 3/8 and finally 2/7. We multiply these together: (1*4*3*2)/(2*9*8*7)=24/1008. This is the probability that every single person in the committee is a woman. One minus that probability is 984/1008=41/42 which is 97.619% Read more >> Options >> http://www.answers.com?initiator=FFANS
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 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.
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.
----------------------------------------------------------------------------------------------------------- ANSWER: Depends on the first supervisor's age and interests. -----------------------------------------------------------------------------------------------------------
The Probability of NOT reading newspaper a is .8 The Probability of NOT reading Newspaper b is .84 The probability of NOT reading Newspaper c is .86 Therefore, .8*.84*.86=0.57792=57.792%
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.
There being 365 days in a year and 50 being less than 365 therefore 2 even far less than that the chances are virtually 0. ______________________ Assume that all 366 days (including leap day) are equally likely to be a person's birthday. The probability that none of them share a birthday is 1*P(second person selected doesn't share a birthday with first person selected)*P(third person selected doesn't share a birthday with first or second person selected)*...*P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected). P(second person selected doesn't share a birthday with first person selected) = 365/366 P(third person selected doesn't share a birthday with first or second person selected)=364/366 . . . P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected)=317/366 P(none share a birthday)=(365/366)*(364/366)*...*(317/366), which is approximately .0299. P(at least two share a birthday) = 1-(365/366)*(364/366)*...*(317/366)=1-.0299=.9701 = 97.01%.
A person that drives a passenger actually drives the car for the owner. This special essential person is called a Chauffeur.
A "Pilot" is a person who drives a plane.
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%