# Probability - sampling without replacement.

• May 17th 2011, 12:14 PM
Gretchy
Probability - sampling without replacement.
I have a feeling it's not as easy as I originally thought!

"There are 20 solar cells, 12 of which are flat and the rest are concentrated. If you were to pick six random cells, what is the probability you'd get three of each?"

So... 60% of all the cells are flat, meaning that the remaining 40% are concentrated and those would also be the probabilities of grabbing all six cells of the same description.

would this formula apply here: (p^x)(1-p)^(x-n)

where p is the probability, x being the description and n being the total cells?

(0.6^12)(1-0.6)^(12-20) = (0.002177) ÷ (0.000655) = 3.324 ??

A little lost, as you can surely see!

Thank you!
• May 17th 2011, 12:24 PM
Plato
Quote:

Originally Posted by Gretchy
I have a feeling it's not as easy as I originally thought!
"There are 20 solar cells, 12 of which are flat and the rest are concentrated. If you were to pick six random cells, what is the probability you'd get three of each?"

There are $\binom{12}{3}\binom{8}{3}$ ways to select three of each.

How many ways to select twelve altogether?

What do you do with that?
• May 17th 2011, 01:10 PM
Gretchy
Quote:

Originally Posted by Plato
There are $\binom{12}{3}\binom{8}{3}$ ways to select three of each.

How many ways to select twelve altogether?

What do you do with that?

Hmm.

I know that $\binom{n}{x}p^x(1-p)^{n-x}$ where $\binom{n}{x} = \frac{n!}{x!(n-x)!}$, then

$\binom{12}{3} = \frac{12!}{3!(9!)} = 220$

Plug it in and...

$220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]$

$= 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}]$
$= 220\cdot [(0.6)^{3}\cdot(0.4)^{9}]$
$= 220\cdot [(0.216)\cdot(0.000262)]$
$= 220 \cdot (0.000056592)$
$= 0.01245024$

Similarly,
$\binom{8}{3} = \frac{8!}{3!(5!)} = 56$

$56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}]$

$= 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]$
$= 56\cdot [(0.064)\cdot(0.7776)]$
$= 56\cdot(0.0497664)$
$= 2.7869184$

So... I add them? $(3 + 3 = 6)$

$2.7869184 + 0.01245024 = 2.79936864$

Yes? No?

(PS: LaTex is NOT easy...!]
• May 17th 2011, 01:21 PM
Plato
Quote:

Originally Posted by Gretchy
Hmm.

I know that $\binom{n}{x}p^x(1-p)^{n-x}$ where $\binom{n}{x} = \frac{n!}{x!(n-x)!}$, then

$\binom{12}{3} = \frac{12!}{3!(9!)} = 220$

Plug it in and...

$220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]$

$= 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}]$
$= 220\cdot [(0.6)^{3}\cdot(0.4)^{9}]$
$= 220\cdot [(0.216)\cdot(0.000262)]$
$= 220 \cdot (0.000056592)$
$= 0.01245024$

Similarly,
$\binom{8}{3} = \frac{8!}{3!(5!)} = 56$

$56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}]$

$= 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]$
$= 56\cdot [(0.064)\cdot(0.7776)]$
$= 56\cdot(0.0497664)$
$= 2.7869184$

So... I add them? $(3 + 3 = 6)$

$2.7869184 + 0.01245024 = 2.79936864$
Yes? No?

SORRY, BUT NO.
All of that is a serious misunderstanding of this problem.
You see this as a binomial distribution. It is not.
In binomial distributions the trials must be independent. These are not.
This sampling without replacement, so not independent.
• May 17th 2011, 01:46 PM
SpringFan25
thats my fault i told him it was binomial in my (removed) post. sorry gretchy!
The text passage contains a probability question related to sampling without replacement, and the user attempts to solve it but uses the wrong approach. The mentor (Plato) corrects the user and provides the correct method to solve the problem.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Probability - sampling without replacement.

• May 17th 2011, 12:14 PM
Gretchy
Probability - sampling without replacement.
I have a feeling it's not as easy as I originally thought!

"There are 20 solar cells, 12 of which are flat and the rest are concentrated. If you were to pick six random cells, what is the probability you'd get three of each?"

So... 60% of all the cells are flat, meaning that the remaining 40% are concentrated and those would also be the probabilities of grabbing all six cells of the same description.

would this formula apply here: (p^x)(1-p)^(x-n)

where p is the probability, x being the description and n being the total cells?

(0.6^12)(1-0.6)^(12-20) = (0.002177) ÷ (0.000655) = 3.324 ??

A little lost, as you can surely see!

Thank you!
• May 17th 2011, 12:24 PM
Plato
Quote:

Originally Posted by Gretchy
I have a feeling it's not as easy as I originally thought!
"There are 20 solar cells, 12 of which are flat and the rest are concentrated. If you were to pick six random cells, what is the probability you'd get three of each?"

There are $\binom{12}{3}\binom{8}{3}$ ways to select three of each.

How many ways to select twelve altogether?

What do you do with that?
• May 17th 2011, 01:10 PM
Gretchy
Quote:

Originally Posted by Plato
There are $\binom{12}{3}\binom{8}{3}$ ways to select three of each.

How many ways to select twelve altogether?

What do you do with that?

Hmm.

I know that $\binom{n}{x}p^x(1-p)^{n-x}$ where $\binom{n}{x} = \frac{n!}{x!(n-x)!}$, then

$\binom{12}{3} = \frac{12!}{3!(9!)} = 220$

Plug it in and...

$220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]$

$= 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}]$
$= 220\cdot [(0.6)^{3}\cdot(0.4)^{9}]$
$= 220\cdot [(0.216)\cdot(0.000262)]$
$= 220 \cdot (0.000056592)$
$= 0.01245024$

Similarly,
$\binom{8}{3} = \frac{8!}{3!(5!)} = 56$

$56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}]$

$= 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]$
$= 56\cdot [(0.064)\cdot(0.7776)]$
$= 56\cdot(0.0497664)$
$= 2.7869184$

So... I add them? $(3 + 3 = 6)$

$2.7869184 + 0.01245024 = 2.79936864$

Yes? No?

(PS: LaTex is NOT easy...!]
• May 17th 2011, 01:21 PM
Plato
Quote:

Originally Posted by Gretchy
Hmm.

I know that $\binom{n}{x}p^x(1-p)^{n-x}$ where $\binom{n}{x} = \frac{n!}{x!(n-x)!}$, then

$\binom{12}{3} = \frac{12!}{3!(9!)} = 220$

Plug it in and...

$220\cdot [(0.6)^{3}\cdot(1-0.6)^{12-3}]$

$= 220\cdot [(0.6)^{3}\cdot(1-0.6)^{9}]$
$= 220\cdot [(0.6)^{3}\cdot(0.4)^{9}]$
$= 220\cdot [(0.216)\cdot(0.000262)]$
$= 220 \cdot (0.000056592)$
$= 0.01245024$

Similarly,
$\binom{8}{3} = \frac{8!}{3!(5!)} = 56$

$56\cdot [(0.4)^{3}\cdot(1-0.4)^{8-3}]$

$= 56\cdot [(0.4)^{3}\cdot(0.6)^{5}]$
$= 56\cdot [(0.064)\cdot(0.7776)]$
$= 56\cdot(0.0497664)$
$= 2.7869184$

So... I add them? $(3 + 3 = 6)$

$2.7869184 + 0.01245024 = 2.79936864$
Yes? No?

SORRY, BUT NO.
All of that is a serious misunderstanding of this problem.
You see this as a binomial distribution. It is not.
In binomial distributions the trials must be independent. These are not.
This sampling without replacement, so not independent.
• May 17th 2011, 01:46 PM
SpringFan25
thats my fault i told him it was binomial in my (removed) post. sorry gretchy!
user
The text passage contains a probability question related to sampling without replacement, and the user attempts to solve it but uses the wrong approach. The mentor (Plato) corrects the user and provides the correct method to solve the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER