# Math Help - Sampling withOUT replacement.

1. ## Sampling withOUT replacement.

1. "With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. A company has just manufactures 837 CDs, and 339 are defective. If 7 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted?"

-I tried 837-339, then dividing it by 7, but apparently that's not correct.

2. A basket contains 11 eggs, 4 of which are cracked. If we randomly select 5 of the eggs for hard boiling, what is the probability of the following events?
a. All of the cracked eggs are selected.
b. None of the cracked eggs are selected.
c. Two of the cracked eggs are selected.

-I've done this one over and over. Sometimes I get 1 right, sometimes 2, but never all of them. What am I doing wrong?!

2. 1. If n is the total number of CD and m is the nember of CD 'OK' at the first test the probability to pass is . If first test is passed the probability to pass the second test is . The probability to pass seven consecutive tests without replacement is then...

(1)

Fot m=498 and n=837 is P=.025945...

Kind regards

$\chi$ $\sigma$

3. Hello, stephmhenry!

1. A sample of items is randomly selected without replacement
and the entire batch is accepted if every item in the sample is okay.
A company has just manufactures 837 CDs, and 339 are defective.
If 7 of these CDs are randomly selected for testing,
what is the probability that the entire batch will be accepted?
$\text{There are: }\:_{837}C_7 \:=\:{837\choose7} \:=\:\frac{837!}{7!\,830!} \text{ possible outcomes.}$

$\text{The entire batch is accepted if 7 of the 498 good CDs are selected.}$

. . $\text{There are: }\:_{498}C_7 \:=\:{498\choose7} \:=\:\frac{498!}{7!\,491!}\text{ ways.}$

$\text{Therefore: }\;P(\text{accepted}) \;=\;\dfrac{_{498}C_7}{_{837}C_7}$

2. A basket contains 11 eggs, 4 of which are cracked.
We randomly select 5 of the eggs for hard boiling.
$\text{There are: }\:{11\choose5} \:=\:462\text{ possible outcomes.}$

What is the probability that:

a. All of the cracked eggs are selected.

We want all 4 of the cracked eggs and 1 good egg.

$\text{There are: }\:{4\choose4}{7\choose1} \:=\:7\text{ ways.}$

$\text{Therefore: }\:P(\text{4 cracked}) \;=\;\frac{7}{462} \;=\;\frac{1}{66}$

b. None of the cracked eggs are selected.

We want 5 of the 7 good eggs.

$\text{There are: }\:{7\choose5} \:=\:21\text{ ways.}$

$\text{Therefore: }\:P(\text{no cracked}) \;=\;\frac{21}{462} \;=\;\frac{1}{22}$

c. Two of the cracked eggs are selected.

We want 2 of he 4 cracked eggs and 3 of the 7 good eggs.

$\text{There are: }\:{4\choose2}{7\choose3} \;=\;6\cdot35 \;=\;210\text{ ways.}$

$\text{Therefore: }\:P(\text{2 cracked}) \;=\;\frac{210}{462} \;=\;\frac{5}{11}$