Probability problem

D

damon354

New member
Joined
Dec 23, 2020
Messages
9
Hello,

The problem:

Mike's playlist contains 20 songs, including 3 from the Beatles. Determine the probability that, in random mode, two Beatles songs are played one after the other.


My reasoning:

The number of combinations for the playlist in random mode is 20 factorial, and to determine the number of combinations containing two Beatles songs in a row, I thought that there can be 19 slots for these two songs: in positions 1 and 2, 2 and 3, ... 19 and 20.
Moreover, as there are 3 Beatles songs, there are 6 pairs of 2 songs possible: AB, BA, AC, CA, BC, CB. So there are 19 x 6 = 114 possible playlists containing 2 Beatles songs in a row on a total of 20! possible playlists. So, the probability we are looking for is [MATH]\frac{114}{20!} [/MATH].

Is it correct ?
 
Don't forget that along with the 3 Beatles songs, there are also 17 others to be permuted!
 
damon354 said:
Mike's playlist contains 20 songs, including 3 from the Beatles. Determine the probability that, in random mode, two Beatles songs are played one after the other.
Click to expand...
Dr.Peterson said:
Don't forget that along with the 3 Beatles songs, there are also 17 others to be permuted!
Click to expand...
Also, tell us if the question means at least two together or exactly two.
Assuming that no song is played twice, there are \(18\cdot 17\cdot 16\cdot 17!\) ways to play no two Beatles recordings consecutively.
The complement of that is the number of ways to play at least two Beatles recordings consecutively.
 
What he is saying first is that "random" mode may mean that a song can be played more than once. You have not told us how "random" mode works. You seem to be assuming that "random" mode can NEVER repeat. If that is true, you could never play more than 20 songs in a life time. Sounds sort of stupid. It seems more likely that either repetitions of the same song are permitted, either immediately or after n songs.

You play a song. The probability that it is a Beatles song is 3/20. You play a second song. The probability that it is a Beatles song is either 3/20 or 2/19 depending on the rules for "random" mode. So we either have 9/400 or 6/380 as the probability that two songs played in succession are both Beatles songs.

Now it is possible that the question is asking about the probability of two Beatles songs being played in succession when you play n songs. We still need to know how "random" mode works, but the problem now also depends on whether it means exactly once that they are played in succession or at least once that they are played in succession.

If you quoted the problem exactly and completely (as we request), we would be able to determine whether the question is even answerable.
 
This is a standard question. Usually it is done with a row of people.
Consider a boy's school senior class of twenty students that has a set of triplets. In a random line-up of the class what is the probability that at least two of the triplets are standing together? I suggest approaching this through the complement.
How ways are there for are there to from a row in which none of triplets are standing together?
Well, the seventeen other students could be used to separate the triplets.
The separators create eight places to put the triplets (183)=816.(183)=816. SEE HERE
There are \(17!=
This is a standard question. Usually it is done with a row of people.
Consider a boy's school senior class of twenty students that has a set of triplets. In a random line-up of the class what is the probability that at least two of the triplets are standing together? I suggest approaching this through the complement.
How ways are there for are there to from a row in which none of triplets are standing together?
Well, the seventeen other students could be used to separate the triplets.
The separators create eight places to put the triplets (183)=816.(183)=816. SEE HERE
There are \(17!=355687428096000\) ways to arrange the separators.
So the the probability that is a random row none if the triplets are standing together is \(\dfrac{[18\cdot 17\cdot 16][17]}{20!}=0.7157\)
Thus the answer is \(1-0.7157\) at least two triplets stand together.
 
damon354 said:
The question means at least two together. Can you clarify your explanation?
Click to expand...
I'm interpreting the question as about a shuffle, not a true random algorithm, in line with your evident assumption, and also in order to make it answerable. You should really confirm the interpretation.

So now we are interpreting the entire problem this way:

Mike's playlist contains 20 songs, including 3 from the Beatles. Determine the probability that, if the 20 songs are shuffled and the list is played once, at least two Beatles songs are played one after the other.​

What pka is doing is to use the complement, a very common trick: the probability that at least two are played consecutively is 1 minus the probability that none are played consecutively. To find the latter, we need to count ways to arrange the songs without putting any two of the three together.

I didn't immediately see where he got his calculation, though. It's easy to come up with different ways to count the same thing, and looking at the numbers doesn't always make it clear where they came from. My own approach would be to select three nonconsecutive slots out of 20, then place the three songs into those, and the other 17 into the other slots. The first step could be a little tricky, because, for example, if you've picked one, it may eliminate either one or two other slots from being included.

But you can instead first put the 17 non-Beatles in some order. This leaves 18 places before, after, and between them, in each of which a Beatles song can be placed. This ensures that each will be isolated. How many places are there to put the first Beatles song? The second? The third? How many ways can you put the 17 others into place?

When I do this, it does result in pka's calculation.
 
Yes "shuffle" was the right word. Following Dr. Peterson's and pka's reasoning, the searched probabilyty is 1 - (16x17x18x17!)/20! = 27/95. Thank you.
 
You must log in or register to reply here.
Top