focus group probability question

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GabeFried

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A group of four members is to be randomly selected from an engineer team consisting of five engineers and seven technologist A.) what is the probability that the focus group will be comprised of engineers only? B.) what is the probability that the focus group will be comprised of 2 engineers and 2 technologists? C.) what is the probability that the focus group will have at least 2 technologist?




for A im thinking i would have to do 5C4 due to the 7 technologists are excluded so 5 is the engineers and the 4 is coming from the group size

for b im thinking 5c2 x 7c2 =210 but im unsure how to make that into a %age if that way is correct

and for c im completely lost
thanks for any help
 
GabeFried said:
A group of four members is to be randomly selected from an engineer team consisting of five engineers and seven technologist A.) what is the probability that the focus group will be comprised of engineers only? B.) what is the probability that the focus group will be comprised of 2 engineers and 2 technologists? C.) what is the probability that the focus group will have at least 2 technologist?

for A im thinking i would have to do 5C4 due to the 7 technologists are excluded so 5 is the engineers and the 4 is coming from the group size
Click to expand...
That's right. What is the denominator?

GabeFried said:
for b im thinking 5c2 x 7c2 =210 but im unsure how to make that into a %age if that way is correct

and for c im completely lost
Click to expand...
Again, what is the denominator? That is, how many possible random choices are there ignoring conditions?

For (c), they are asking for the probability that 2, 3, 4, 5, 6, or 7 technologists will be chosen. It will probably be easier to find the probability of 0 or 1 technologists (part of which you already have), and subtract from 1.
 
Dr.Peterson said:
That's right. What is the denominator?


Again, what is the denominator? That is, how many possible random choices are there ignoring conditions?

For (c), they are asking for the probability that 2, 3, 4, 5, 6, or 7 technologists will be chosen. It will probably be easier to find the probability of 0 or 1 technologists (part of which you already have), and subtract from 1.
Click to expand...

and for B would it be 210/48 x100 equaling 22.85% rounded to 22.9%?
 
GabeFried said:
A group of four members is to be randomly selected from an engineer team consisting of five engineers and seven technologist
A.) what is the probability that the focus group will be comprised of engineers only?
B.) what is the probability that the focus group will be comprised of 2 engineers and 2 technologists?
C.) what is the probability that the focus group will have at least 2 technologist? and for c im completely lost
Click to expand...
For part C) I would find the probability of having a committee with no technologist or just one technologist. Subtract from 1.
There are \(\dbinom{5}{4}\cdot\dbinom{7}{0}+\dbinom{5}{3}\cdot\dbinom{7}{1}\) ways the select such a committee,
 
GabeFried said:
denominator would be 48 as 5+7=12 x4 so 10.4% for A
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Are you saying there are only 48 ways to choose a group of 4 from among 12 people? That's what the denominator has to be.
GabeFried said:
and for B would it be 210/48 x100 equaling 22.85% rounded to 22.9%?
Click to expand...
No, you numerator is larger than your denominator, meaning you think there are more ways to choose 2 and 2 than there are ways to choose with no conditions!

And 210/48 = 4.375, not 0.2285. You divided the wrong way, probably because you knew the answer has to be less than 100%.

Once you get the right denominator, things will make sense.
 
Dr.Peterson said:
Are you saying there are only 48 ways to choose a group of 4 from among 12 people? That's what the denominator has to be.

No, you numerator is larger than your denominator, meaning you think there are more ways to choose 2 and 2 than there are ways to choose with no conditions!

And 210/48 = 4.375, not 0.2285. You divided the wrong way, probably because you knew the answer has to be less than 100%.

Once you get the right denominator, things will make sense.
Click to expand...
would it be 12c4=495?
 
pka said:
For part C) I would find the probability of having a committee with no technologist or just one technologist. Subtract from 1.
There are \(\dbinom{5}{4}\cdot\dbinom{7}{0}+\dbinom{5}{3}\cdot\dbinom{7}{1}\) ways the select such a committee,
Click to expand...
thank you!
 
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