Probability of 3 songs

[rachelmaddie]

I need help with the formula for probability.

 

[Deleted member 4993]

I need help with the formula for probability.View attachment 14360

Suppose Noam had to choose ONE song only.

What is the probability that he will choose "The entertainer"?

 

[Steven G]

If you were to pick 3 songs from the pile of 20 songs in how many ways can you pick The Entertainer, Something Doing and Ragtime Dance? Knowing this answer is the start of you having the solution.

 

[rachelmaddie]

I don’t know how to write it out.
Would it be probability of Noam choosing one song is 1/20?

 

[Steven G]

I don’t know how to write it out.
Would it be probability of Noam choosing one song is 1/20?

It really is not that hard! There is only one way to pick The Entertainer, Something Doing and Ragtime Dance if you only pick 3 songs and that one way is to pick The Entertainer, Something Doing and Ragtime Dance. Do you see that? Now how many ways can you pick any 3 songs from 20?

 

[rachelmaddie]

It really is not that hard! There is only one way to pick The Entertainer, Something Doing and Ragtime Dance if you only pick 3 songs and that one way is to pick The Entertainer, Something Doing and Ragtime Dance. Do you see that? Now how many ways can you pick any 3 songs from 20?

1/20 x 1/19 x 1/18

 

[Steven G]

How many ways can you choose 3 songs from 20? You must answer this.

The final answer will be in the form of (# of favorable outcomes)/(# of total outcomes) = 1/(# of total outcomes) = 1/(How many ways can you choose 3 songs from 20)

 

[rachelmaddie]

How many ways can you choose 3 songs from 20? You must answer this.

The final answer will be in the form of (# of favorable outcomes)/(# of total outcomes) = 1/(# of total outcomes) = 1/(How many ways can you choose 3 songs from 20)

So I would not multiply the three values?

 

[pka]

Do you understand combinations? That \(\displaystyle \mathcal{C}^{20}_3=\dbinom{20}{3}=\dfrac{20!}{3!\cdot(20-3)!}\) that is the number of the combinations of twenty songs choosing three.
Since you specify three particular songs that is only one event so that its probability is \(\displaystyle \dfrac{{\large 1}}{\binom{20}{3}}\)

 

[rachelmaddie]

Do you understand combinations? That \(\displaystyle \mathcal{C}^{20}_3=\dbinom{20}{3}=\dfrac{20!}{3!\cdot(20-3)!}\) that is the number of the combinations of twenty songs choosing three.
Since you specify three particular songs that is only one event so that its probability is \(\displaystyle \dfrac{{\large 1}}{\binom{20}{3}}\)

Is there another method to do this?

 

[Deleted member 4993]

Is there another method to do this?

Not really!

You need to think this situation through. Think about it by choosing 3 records from 4 record collection.

Then work up your way.

 

[rachelmaddie]

Not really!

You need to think this situation through. Think about it by choosing 3 records from 4 record collection.

Then work up your way.

I need more help understanding the formula used here

 

[rachelmaddie]

I’ve found some resources. Is this the method I should follow??

 

[rachelmaddie]

I’ve attempted the first part.
20C3 = 20!/(20-3)! (3)!

 

[pka]

I’ve found some resources. Is this the method I should follow??View attachment 14374

I really don't know the answer to your question. It is exactly the method I used in reply #9.

 

[rachelmaddie]

I really don't know the answer to your question. It is exactly the method I used in reply #9.

My question is how do I get the solution like in the picture?

 

[pka]

I’ve attempted the first part.
20C3 = 20!/(20-3)! (3)!

If I were you, I would learn to use WolfFramAlpha See here

 

[rachelmaddie]

If I were you, I would learn to use WolfFramAlpha See here

1140?

 

[rachelmaddie]

1140?

1/1140?

 

[pka]

My question is how do I get the solution like in the picture?View attachment 14375

Use the link I just posted. You can change \(\displaystyle 20\to 15~\&~3\to 9\)
You can change them right there.