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You select 12 people from a group of 30 people that includes 12 men and 18 women what is the probability that the group will be 12 women?
The probability that 12 randomly selected people from a group of 12 men and 18 women will all be women is (18 in 30) times (17 in 29) times (16 in 28) times (15 in 27) times (14 in 26) times (13 in 25) times (12 in 24) times (11 in 23) times (10 in 22) times (9 in 21) times (8 in 20) times (7 in 19), which is equal to 8,892,185,702,400 in 2,180,547,008,640,000, which is equal to 68 in 16,675,which is equal to 0.00407796.
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What is the probabililty of at least 2 people same birthday from a group of 13 people?
19.4%CALCULATION:The probability of at least 2 people having the same birthday in a group of 13people is equal to one minus the probability of non of the 13 people having thesame birthday.Now, lets estimate the probability of non of the 13 people having the same birthday.(We will not consider 'leap year' for simplicity, plus it's effect on result is minimum)1. We select the 1st person. Good!.2. We select the 2nd person. The probability that he doesn't share the samebirthday with the 1st person is: 364/365.3. We select the 3rd person. The probability that he doesn't share the samebirthday with 1st and 2nd persons given that the 1st and 2nd don't share the samebirthday is: 363/365.4. And so forth until we select the 13th person. The probability that he doesn'tshare birthday with the previous 12 persons given that they also don't sharebirthdays among them is: 353/365.5. Then the probability that non of the 13 people share birthdays is:P(non of 13 share bd) = (364/365)(363/365)(362/365)∙∙∙(354/365)(353/365)P(non of 13 share bd) ≈ 0.805589724...Finally, the probability that at least 2 people share a birthday in a group of 13people is ≈ 1 - 0.80558... ≈ 0.194 ≈ 19.4%The above expression can be generalized to give the probability of at least x =2people sharing a birthday in a group of n people as:P(x≥2,n) = 1 - (1/365)n [365!/(365-n)!]
Is a sample really random if only a select group of people respond?
no
How many ways can you select two people from a group of 9 people?
To select two people from a group of 9, you can use the combination formula, which is given by ( C(n, k) = \frac{n!}{k!(n-k)!} ). For this scenario, ( n = 9 ) and ( k = 2 ). Therefore, the number of ways to select two people is ( C(9, 2) = \frac{9!}{2!(9-2)!} = \frac{9 \times 8}{2 \times 1} = 36 ). Thus, there are 36 ways to select two people from a group of 9.
What is the probability of you and 2 complete random people sharing a birthday?
Birthdays are not uniformly distributed over the year. Also, if you were born on 29 February, for example, the probability would be much smaller. Ignoring these two factors, the probability is 0.0082
A group's reported are more likely to become equal to the statistical probability of loss?
the larger the group, the more likely the statistical probability of loss will be equal
What is the probabililty of at least 2 people same birthday from a group of 13 people?
19.4%CALCULATION:The probability of at least 2 people having the same birthday in a group of 13people is equal to one minus the probability of non of the 13 people having thesame birthday.Now, lets estimate the probability of non of the 13 people having the same birthday.(We will not consider 'leap year' for simplicity, plus it's effect on result is minimum)1. We select the 1st person. Good!.2. We select the 2nd person. The probability that he doesn't share the samebirthday with the 1st person is: 364/365.3. We select the 3rd person. The probability that he doesn't share the samebirthday with 1st and 2nd persons given that the 1st and 2nd don't share the samebirthday is: 363/365.4. And so forth until we select the 13th person. The probability that he doesn'tshare birthday with the previous 12 persons given that they also don't sharebirthdays among them is: 353/365.5. Then the probability that non of the 13 people share birthdays is:P(non of 13 share bd) = (364/365)(363/365)(362/365)∙∙∙(354/365)(353/365)P(non of 13 share bd) ≈ 0.805589724...Finally, the probability that at least 2 people share a birthday in a group of 13people is ≈ 1 - 0.80558... ≈ 0.194 ≈ 19.4%The above expression can be generalized to give the probability of at least x =2people sharing a birthday in a group of n people as:P(x≥2,n) = 1 - (1/365)n [365!/(365-n)!]
What is the probability of 2 or more people in a a group of about 30 having the same birthday?
The probability with 30 people is 0.7063 approx.
What is another way to say a group of people?
a select group
Lamone is in a group of 10 students on waiting list for a school trip There is space for 2 more students on the tripso Mr.Baraka plansto randomly select 2 names from a hatWhat is the probability?
The answer depends on what is the probability of WHAT!
Which select group of people is Christmas celebrated by?
christians
Who are they going tobenefit from it?
Only a select group of people.
A group of like-minded people who select candidates to support in an upcoming election is known as a .?
Caucus
Is a sample really random if only a select group of people respond?
no
What is the probabililty of at least 2 people same birthday from a group of 12 people?
The probability of at least 2 people in a group of npeople sharing a common birthday can be expressed more easily (mathematically) as 1 minus the probability that nobody in the group shares a birthday. Consider two people. The probability that they don't have a common birthday is 365/365 x 364/365. So the probability that they do share a birthday is 1-(365/365 x 364/365) = 1-365x364/3652 Now consider 3 people. The probability that at least 2 share a common birthday is 1-365x364x363/3653 And so on so that the probability that at least 2 people in a group of n people having the same birthday = 1-(365x363x363x...x365-n+1)/365n = 1-365!/[ (365-n)! x 365n ]In the case of 12 people this equates to 0.16702 (or 16.7%).
What is the probability that at least 2 people have the same birth month in a group of 8 people?
Birth months are not uniformly distributed across the year. However, if yo assume that they are, the probability is 0.9536 (approx).
What is a group of people who help to select the president and the vice president?
The Electoral College
What is the probability of randomly selecting a sick person from a group of three sick people and seven healthy people?
3/10 or 0.3
