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)
The concept of the "chosen people" typically refers to the Jewish people in the context of Judaism. Other cultures and societies have their own historical and religious narratives that shape their beliefs and identities. It is important to understand and respect the diversity of beliefs and perspectives across different cultures.