EX 2.22 Beatles' songs

• A particular iPod playlist contains 100 songs,

  • 10 of which are by the Beatles.

• Suppose the shuffle feature is used to play the songs in random order

  • the randomness of the shuffling process is investigated in “Does Your iPod Really Play Favorites?” (The Amer. Statistician, 2009: 263-268).

• What is the probability that the first Beatles song heard is the fifth song played?

• In order for this event to occur, it must be the case that

  • the first four songs played are not Beatles’ songs (NBs)
  • the fifth song is by the Beatles (B).

• The number of ways to select the first five songs is $100\left(99\right)\left(98\right)\left(97\right)\left(96\right).$

• The number of ways to select these five songs so that

  • the first four are NBs
  • the next is a B is 90(89)(88)(87)(10).

• The random shuffle assumption implies that any particular set of 5 songs from amongst the 100 has the same chance of being selected as the first five played as does any other set of five songs;

  • each outcome is equally likely.

• Therefore the desired probability is the ratio of the number of outcomes for which the event of interest occurs to the number of possible outcomes:

$$\begin{array}{rl}& P\left(1^{\text{st }}\mathrm{B}\text{ is the }{5}^{\text{th }}\text{ song played }\right)\\ =& \frac{90\cdot 89\cdot 88\cdot 87\cdot 10}{100\cdot 99\cdot 98\cdot 97\cdot 96}\\ =& \frac{P_{4,90}\cdot \left(10\right)}{P_{5,100}}\\ =& .0679\end{array}$$

Here is an alternative line of reasoning involving combinations.

• Rather than focusing on selecting just the first five songs,

  • think of playing all 100 songs in random order.

• The number of ways of choosing 10 of these songs to be the Bs is $\left(\begin{array}{c}100\\ 10\end{array}\right)$

  • without regard to the order in which they are then played

• Now if we choose 9 of the last 95 songs to be Bs,

  • which can be done in $\left(\begin{array}{c}95\\ 9\end{array}\right)$ ways,
  • that leaves four NBs and one $\mathrm{B}$ for the first five songs.

• There is only one further way for these five to

  • start with four NBs and
  • then follow with a B
  • remember that we are considering unordered subsets

• Thus $P\left(1^{\text{st }}\mathrm{B}\text{ is the }{5}^{\text{th }}\text{ song played }\right)=\frac{\left(\begin{array}{c}95\\ 9\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}$

  • It is easily verified that this latter expression is in fact identical to the first expression for the desired probability,
    • so the numerical result is again .0679.

• The probability that one of the first five songs played is a Beatles’ song is

$$\begin{array}{rl}& P\left(1^{\text{st}}\mathrm{B}\text{ is the }{1}^{\text{st}}\text{ or }{2}^{\text{nd}}\text{ or }{3}^{\text{rd}}\text{ or }{4}^{\text{th}}\text{ or }{5}^{\text{th}}\text{ song played }\right)\\ & =\frac{\left(\begin{array}{c}99\\ 9\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}+\frac{\left(\begin{array}{c}98\\ 9\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}+\frac{\left(\begin{array}{c}97\\ 9\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}+\frac{\left(\begin{array}{c}96\\ 9\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}+\frac{\left(\begin{array}{c}95\\ 9\end{array}\right)}{\left(\begin{array}{c}100\\ 10\end{array}\right)}\\ & =.4162\end{array}$$

• It is thus rather likely that a Beatles’ song will be one of the first five songs played.
• Such a “coincidence” is not as surprising as might first appear to be the case.