Permutations/combinations: If 3 of 10 perform farcical comedy, what is prob. that exactly 2 of random 4 do farcical?

[Audentes]

Hi, I have permutations and combinations problems to do on IXL and I honestly don't know how to do them. I've read the explanations after getting some wrong & it doesn't make sense to me.
Here is one of the problems:

A comedy club hosts an amateur comedy night, a good opportunity for aspiring comedians to perform. Of the 10 amateur comedians who signed up for tonight's show, 3 perform farcical comedy. If the host randomly selects 4 comedians from the sign-up list, what is the probability that exactly 2 of the chosen comedians practice farcical? Write your answer as a decimal rounded to four decimal places.

I want to say that the answer is 4c2/10c3 but I have a feeling that is wrong because I fhought of that answer the same way I did for the problems I got wrong.

could someone help me out?


thanks
FR

 

[R.M.]

Start with these questions: What is the probability that 1 person performs farcical comedy? How many times do you need that probability? How many ways can those people be chosen?

 

[khansaheb]

How many ways can you choose exactly [1 farcical + 2 other comedian] ?

Now you will choose 1 exactly 1 more farcical comedian out of the rest?

 

[Dr.Peterson]

A comedy club hosts an amateur comedy night, a good opportunity for aspiring comedians to perform. Of the 10 amateur comedians who signed up for tonight's show, 3 perform farcical comedy. If the host randomly selects 4 comedians from the sign-up list, what is the probability that exactly 2 of the chosen comedians practice farcical? Write your answer as a decimal rounded to four decimal places.

I want to say that the answer is 4c2/10c3 but I have a feeling that is wrong because I fhought of that answer the same way I did for the problems I got wrong.

We could help better if you told us your thinking, not just your answer. What does your numerator mean to you?

We have 10 (distinguishable) people FFFOOOOOOO, and want to choose FFOO (ignoring order, but distinguishing people).

I'd start by asking, How many ways are there to choose 2 Farcicals, and 2 Others?

 

[Audentes]

Start with these questions: What is the probability that 1 person performs farcical comedy? How many times do you need that probability? How many ways can those people be chosen?

3/10 and that probability is needed twice?

How many ways can you choose exactly [1 farcical + 2 other comedian] ?

Now you will choose 1 exactly 1 more farcical comedian out of the rest?

252 ways? (3 times 2 times 7 times 6?)

We could help better if you told us your thinking, not just your answer. What does your numerator mean to you?

We have 10 (distinguishable) people FFFOOOOOOO, and want to choose FFOO (ignoring order, but distinguishing people).

I'd start by asking, How many ways are there to choose 2 Farcicals, and 2 Others?

252?

 

[Dr.Peterson]

3/10 and that probability is needed twice?

252 ways? (3 times 2 times 7 times 6?)

252?

I didn't ask you to give a numerical answer for that question; I asked you to explain your thinking (in words), and suggested that as a question to ask yourself. What did you do to get that number, and why? Then, how might it be used to solve the problem?

 

[Audentes]

I didn't ask you to give a numerical answer for that question; I asked you to explain your thinking (in words), and suggested that as a question to ask yourself. What did you do to get that number, and why? Then, how might it be used to solve the problem?

(3c2 * 7c2) / (10c4)?
(population comb. sample * population comb. Sample) / (totalpopulation comb. totalsample)?

 

[Steven G]

You should look at hypergeometric problems. It makes this stuff easy.

Here is an example. You have 10 lightbulbs where exactly three are defective.
What is the probability of choosing 4 light bulbs and getting exactly 2 defective.

Hypergeometric separates this into categories. Category 1 is non-defective bulbs and Category 2 is defective bulbs.
You want to choose 2 defective bulbs from the 3 defective bulbs and 2 non-defective bulbs from the (10-3=) 7 non-defective. You are also choosing 4 bulbs from 10 bulbs.

The answer is [ 4c3*7c2]/10c4

 

[Dr.Peterson]

(3c2 * 7c2) / (10c4)?
(population comb. sample * population comb. Sample) / (totalpopulation comb. totalsample)?

That sounds good; but what I mean by explaining is something more like this:

The numerator is the number of ways to choose 2 of the 3 Farcicals, and 2 of the 7 Others. This is 3C2 * 7C2 = 3 * 21 = 63.

The denominator is the number of ways to choose 4 of 10 people. This is 10C4 = 210.

Therefore the probability is 63/210 = 3/10.

 

[Audentes]

You should look at hypergeometric problems. It makes this stuff easy.

Here is an example. You have 10 lightbulbs where exactly three are defective.
What is the probability of choosing 4 light bulbs and getting exactly 2 defective.

Hypergeometric separates this into categories. Category 1 is non-defective bulbs and Category 2 is defective bulbs.
You want to choose 2 defective bulbs from the 3 defective bulbs and 2 non-defective bulbs from the (10-3=) 7 non-defective. You are also choosing 4 bulbs from 10 bulbs.

The answer is [ 4c3*7c2]/10c4

Thank you for the suggestion!

That sounds good; but what I mean by explaining is something more like this:

The numerator is the number of ways to choose 2 of the 3 Farcicals, and 2 of the 7 Others. This is 3C2 * 7C2 = 3 * 21 = 63.

The denominator is the number of ways to choose 4 of 10 people. This is 10C4 = 210.

Therefore the probability is 63/210 = 3/10.

Thank you!