For what natural values of n is (2-2n)-(5n-27) positive?

1 answer:

8 0

First simplify.
2-2n-5n+27= -7n+29.
then require this to be greater than 0 .
-7n+29>0
from here you can easily get that
n <29/7

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Incomplete question, but I gave a primer on the hypergeometric distribution, which is used to solve this question, so just the formula has to be applied to find the desired probabilities.

Step-by-step explanation:

The resistors 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.

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$C_{n,x}$ 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:

12 resistors, which means that $N = 12$

3 defective, which means that $k = 3$

4 are selected, which means that $n = 4$

To find an specific probability, that is, of x defectives:

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

$P(X = x) = h(x,12,4,3) = \frac{C_{3,x}*C_{9,4-x}}{C_{12,4}}$

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