The probability is (4/5)3 = 64/125
We'd like to know the likelihood that any 5 people chosen from 20 will be left-handed if the probability of being left-handed is assumed to be 10%. Well, 5 people can be chosen from 20 in (20 C 5) = 15504 ways. So, we have (15504)*(.1^5)*(.9^15) = 3.19%. This is a Binomial probability problem, where we are interested in 'k' successes (k = 5, success --> people being left-handed) in 'n' trials (n = 20).
7 women in a group of 11 people 7/11
The probability with 30 people is 0.7063 approx.
Leaving aside leap years, the probability is 0.0137
The probability is approx 0.81
The answer depends on how many are chosen at a time.
The Maracanã Stadium It was also the biggest stadium in the world... and since nowadays, it has the record of people inside a soccer stadium, 195.513 people watching to the match in it.
its around 80,000 to 90,000 people
It depends on how many people enter
A large group of people standing together in one place. Such as in a sports stadium watching football.
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
There are 11 people total and 7 women. The probability the chairman is a woman is 7/11.
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.
We'd like to know the likelihood that any 5 people chosen from 20 will be left-handed if the probability of being left-handed is assumed to be 10%. Well, 5 people can be chosen from 20 in (20 C 5) = 15504 ways. So, we have (15504)*(.1^5)*(.9^15) = 3.19%. This is a Binomial probability problem, where we are interested in 'k' successes (k = 5, success --> people being left-handed) in 'n' trials (n = 20).
people who are chosen people who are chosen people who are chosen
About 18.5 percent thst both will have seatbelts on. (.43 x 2)
27,598 people can fit into Britannia Stadium; the stadium of Stoke City FC.