At a city council meeting, there 101 people present,

Ответ:

abomb6292

25.06.2021

23.63% chance that exactly three is elderly

Step-by-step explanation:

The people are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

101 total people means that N = 101

Sample of 6 means that n = 6

66 elderly means that k = 66

If you talked to six random people there, what is the percent chance that exactly three is elderly?

This is P(X = 3). So

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,101,6,66) = \frac{C_{66,3}*C_{35,3}}{C_{101,6}} = 0.2363$

0.2363*100% = 23.63%

23.63% chance that exactly three is elderly

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