Probability of 3 songs

[rachelmaddie]

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

Is that not correct?

 

[rachelmaddie]

Is that not correct?

That’s the result it gave me.

 

[rachelmaddie]

1/the total number of songs
n represents total number of items
r represents number of items being chosen at a time
n = 20
r = 3
20C3 = 20!/(20-3)! (3)! = 1140
Number of favorable outcomes/total number of outcomes = 1/1140

 

[pka]

1/the total number of songs
n represents total number of items
r represents number of items being chosen at a time
n = 20
r = 3
20C3 = 20!/(20-3)! (3)! = 1140
Number of favorable outcomes/total number of outcomes = 1/1140

CORRECT!

 

[Steven G]

The number of ways to choose r objects from n objects is nCr = n!/(r!(n-r)!)

 

[rachelmaddie]

The number of ways to choose r objects from n objects is nCr = n!/(r!(n-r)!)

That is the formula used

 

[Steven G]

That is the formula used

Yes, good job.
Do you understand why 1 is in the numerator of the answer?

 

[JeffM]

How many ways can I choose 3 from 20

[MATH]\dbinom{20}{3} = \dfrac{20!)(3! * (20 - 3)!} = \dfrac{20 * 19 * 18 * 17!}{3! * 17!} = \dfrac{20 * 19 * 18}{3 * 2} = 1140.[/MATH]
That is the total number of collections of three we can assemble. Exactly one of them is what we want. If all are equally likely (an assumption that is not necessarily warranted), then the probability of getting that selection is indeed

1/1140.

You can get the same answer a different way. Label the three records we want as A, B, and C. The probability of choosing A, B, and C in that order is

[MATH]\dfrac{1}{20} * \dfrac{1}{19} * \dfrac{1}{18} = \dfrac{1}{6840}.[/MATH]
But you do not care in what order you select them. ABC, ACB, BAC, BCA, CAB, and CBA all get you the same collection. Moreover all 6 orders have the same probability. So

[MATH]6 * \dfrac{1}{6840} = \dfrac{1}{1140}.[/MATH]
There are frequently several ways to think about a probability.

 

[firemath]

This is so confusing.