There're 10 songs in an MP3 player. Each time the player plays 5 random songs.

  1. What is the probability that all 5 songs are the same?
  2. What is the probability that at least 1 song will be played more than once?
  3. What is the probability that the first and the last song is "Bohemian Rhapsody by Queen" (it's one of the 10 songs)?
  4. What is the probability that exactly 2 songs will different out of the 5 songs to be played?
  5. What is the probability that one song will be player 3 times and the rest of the songs will not be played more than once?

I know it's a lot of questions but I guess it didn't make sense to open them as separate as they all refer to the same conditions. Here're my answers and logic, not sure if I'm right.

Total amount of possibilities without any restrictions is $10^5$. Let it be $T$.

  1. There's 10 ways that all songs are the same.
  2. First, let's calculate how many ways there're to choose that all songs will be played only once: $P(10,5)$. Let this event be $A$. Hence the restriction in point 2 is essentially the complement of $A$ hence the probability is $1 - A^c = 1 - \frac{P(10,5)}{10^5}$.
  3. Given that the first and the last song was chosen as "Bohemian Rhapsody" then we have 3 songs in between to permute. Which is $9^3$.
  4. First let's choose these 2 different songs: $P(10,2)$. The other 3 songs can be either of the previously chosen 2 songs hence the end result is $P(10,2) * 2^3$.
  5. First we chose the songs to be played 3 times which is $\binom{10}{1}$. The rest can be chose as $P(10,2)$. Hence the result is $\binom{10}{1} * P(10,2)$.

I agree with your answers for #1 and #2.

The way I read #3, there is no restriction against Bohemian Rhapsody playing as song 2, 3 or 4 also. Therefore, I would have used $10^3$ instead or just ignored the three middle songs and calculated the probability directly as 0.1 * 0.1. Either gets the same answer.

For #4, I think your answer imposes the condition that the first 2 songs are not the same which I don't believe is a requirement. Two songs can be played $2^5$ ways. We need to subtract 2 to represent playing all one song or all the other which don't fulfill the condition. Then we have to choose the 2 songs so I get: ${10 \choose 2} * ($2^5 -2)

For #5, your answer doesn't consider that the triple play and the other 2 can be in any order within the 5. Your use of P to choose the 2 single plays already takes their ordering into account, so I would just multiply your answer by ${5 \choose 3}$

  • $\begingroup$ Thanks! You're right regarding #3 and I guess regarding #4, but I'm not sure about the last one, I still have doubts. $\endgroup$ – Yos Dec 7 '16 at 7:26
  • $\begingroup$ I think you are right, #5 is not correct. Better to start over. Choose the trip $\endgroup$ – Tony Mulieri Dec 7 '16 at 15:59
  • $\begingroup$ Sorry. Hit enter too soon. starting over: choose the triple play first is $ 10 \choose 1 $ Then choose the 2 single plays, so $ 9 \choose 2 $ Then the ways they can be arranged is 5!/3! The answer it the product of those three terms (divided by T for probability) $\endgroup$ – Tony Mulieri Dec 7 '16 at 16:04
  • $\begingroup$ I posted the 5th question separately here: math.stackexchange.com/questions/2059626/… You were pretty close $\endgroup$ – Yos Dec 15 '16 at 10:15
  • $\begingroup$ Good writeups. The answer given there is actually the same as my answer in the comment above - 10 choose 1 multiplied by 9 choose 2 is the same as 10 choose 3 multiplied 3 choose 1. They work out he same if you expand them and from a reality point of view, selecting 1 from ten and then 2 from the remaining 9 is the same as selecting 3 from 10 and then picking one from those 3. $\endgroup$ – Tony Mulieri Dec 16 '16 at 15:06

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