2
$\begingroup$

My question is the same as this one, but I did not find either of the answers to it sufficient and do not have enough reputation to comment.

We do not understand part of the example given in Devore's Probability & Statistics textbook. The example problem is set up as follows:

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

The problem is first solved this way, which does make sense to me:

In order for this event to occur, it must be the case that the first four songs played are not Beatles’ songs (NBs) and that the fifth song is by the Beatles (B). The number of ways to select the first five songs is 100(99)(98)(97)(96). The number of ways to select these five songs so that the first four are NBs and 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:

P(1st B is the 5th song played) = $\frac{90 \cdot 89 \cdot 88 \cdot 87 \cdot 10}{100 \cdot 99 \cdot 98 \cdot 97 \cdot 96} = \frac{P_{4, 90} \cdot (10)}{P_{5, 100}} = .0679$

Devore goes on, however, to use a different approach based on combinations (rather than permutations), and it is this part which I fail to understand:

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 (without regard to the order in which they are then played) is $\binom{100}{10}$. Now if we choose 9 of the last 95 songs to be Bs, which can be done in $\binom{95}{9}$ ways, that leaves four NBs and one 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(1st B is the 5th song played) = $\frac{\binom{95}{9}}{\binom{100}{10}}$

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.

$\binom{95}{9}$, according Devore, is "the ways to choose 9 of the last 95 songs to be Bs". Meanwhile, the first answer says that this is "the number of ways [we] can have [the] Beatles play fifth", and the second answer says it is "the number of ways to choose the position of the other 9 B's out of the remaining 95 positions". These three explanations all seem to be different and none of them make sense to me.

To me, it seems that if you have 95 songs left and 9 of them are Beatles songs, there is only one way to choose exactly those 9 Beatles songs out of the total 95 available, so wouldn't the probability of that happening be 1 / $\binom{95}{9}$? And why is the third answerer talking about the positions of the 9 remaining B's? I thought combinations by definition did not take order into consideration.

Regarding the $\binom{100}{10}$ in the denominator, according to Devore it is "the number of ways of choosing 10 of these songs to be the Bs". But again, to me it seems like $\binom{100}{10}$ would be the number of ways you could choose any ten songs out of the hundred - and wouldn't only one of those selections be exactly those ten Beatles songs, for a probability of 1 / $\binom{100}{10}$?

I guess $\require{cancel}\frac{\frac{\cancel{1}}{\binom{95}{9}}}{\frac{\cancel{1}}{\binom{100}{10}}} = \frac{\binom{95}{9}}{\binom{100}{10}}$

... but if this step is being skipped over, it doesn't seem consistent with any of the explanations.

Meanwhile, the two answers to the duplicate of this question respectively state that $\binom{100}{10}$ represents "the total number of ways can play ten beatles songs in 100" and "the number of ways to select the position of all 10 B's out of all 100 positions, which is the universe under consideration". This first answer seems consistent with Devore, whose logic I explained my objection to in the last paragraph. As for the second, I again do not understand it speaks of about position, when combinations do not consider order. And I'm not sure about the part which says it represents "the universe under consideration"... Wouldn't the universe under consideration instead be the five songs played ($\binom{10}{5}$?

Finally, I found this answer at Brainly, which makes complete sense to me but comes up with a totally different answer:

The 1st Beatles song heard is the 5th song played means that first 4 songs are not Beatles songs (there are 100 - 10 = 90 such songs) and the 5th song is Beatles song (there are 10 such songs). Hence, the probability that the first Beatles song heard is the fifth song played is $\frac{\binom{90}{4} \cdot \binom{10}{1}}{\binom{100}{5}} \approx .339$

Is this answer wrong? If so, why?

If not, does that mean Devore is wrong? If Devore is wrong, what is wrong with his first approach?

It's driving me crazy, since Devore's first approach makes total sense to me but results in the same probability as his second approach, which makes no sense to me, yet Devore's probability differs from the Brainly approach, which also makes sense to me.

Please help!

$\endgroup$
1
  • $\begingroup$ Note that $$\frac{\frac{1}{\binom{95}{9}}}{\frac{1}{\binom{100}{10}}} = \frac{\binom{100}{10}}{\binom{95}{9}} > 1$$ $\endgroup$ Dec 7 '19 at 11:01
1
$\begingroup$

Permutations count ordered selections of distinct objects. Combinations count unordered selections of distinct objects. To put it another way, combinations count subsets.

Combinations can be used to count ordered selections when the objects are not distinct. For instance, there are ten sequences of five coin tosses with exactly three heads:

HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, TTHHH

They correspond to the $$\binom{5}{3}$$ ways we can select a subset of three of the five positions in which to place the three heads. The two tails must be placed in the remaining two positions of the sequence.

The answer on Brainly is incorrect.

The probability that the first Beatles song appears in the fifth position can be found by multiplying the probability that no Beatles song appears in the first four positions by the probability that a Beatles song then appears in the fifth position given that no Beatles song has appeared before then. For this to occur, four of the $90$ non-Beatles songs on the playlist must be selected from the $100$ songs on the playlist, then one of the $10$ Beatles songs must be selected from the remaining $96$ songs on the playlist. Thus, the probability that the first Beatles song appears in the fifth position is $$\frac{\dbinom{90}{4}}{\dbinom{100}{4}} \cdot \frac{\dbinom{10}{1}}{\dbinom{96}{1}} \approx 0.0678781822$$ as Devore found.

Why?

Observe that \begin{align*} \frac{90 \cdot 89 \cdot 88 \cdot 87 \cdot 10}{100 \cdot 99 \cdot 98 \cdot 97 \cdot 96} & = \frac{90 \cdot 89 \cdot 88 \cdot 87}{100 \cdot 99 \cdot 98 \cdot 97} \cdot \frac{10}{96}\\ & = \frac{P(90, 4)}{P(100, 4)} \cdot \frac{P(10,1)}{P(96,1)}\\ & = \frac{\dbinom{90}{4}4!}{\dbinom{100}{4}4!} \cdot \frac{\dbinom{10}{1}1!}{\dbinom{96}{1}1!}\\ & = \frac{\dbinom{90}{4}}{\dbinom{100}{4}} \cdot \frac{\dbinom{10}{1}}{\dbinom{96}{1}} \end{align*}

The answer on Brainly is incorrect since it calculates the probability that exactly one of the first five songs selected is a Beatles song rather than the probability that the fifth song selected is the first Beatles song. By not taking the order in which the five songs appear into account, it allows for the first Beatles song to appear in any of the first five positions, which is why the answer on Brainly is five times the correct answer.

To see why Devore's second answer is correct, let's add a few details. We will consider all $100$ positions. We know that exactly ten of these positions will be filled with Beatles songs. If the first Beatles song appears in the fifth position, then none of the first four positions will be filled with a Beatles song, the fifth position will be filled with a Beatles song, and nine of the remaining $95$ positions will be filled with Beatles songs. Hence, the probability that the first Beatles song appears in the fifth position is $$\frac{\dbinom{4}{0}\dbinom{1}{1}\dbinom{95}{9}}{\dbinom{100}{10}} = \frac{\dbinom{95}{9}}{\dbinom{100}{10}}$$ Notice that the numerator counts the number of ways none of the Beatles songs appear in the first four positions, one appears in the fifth position, and the other nine appear in the last $95$ positions, while the denominator counts all the ways the ten Beatles songs could appear among the $100$ positions.

$\endgroup$
3
  • $\begingroup$ Thank you for taking the time to help me! I think I finally understand. :D $\endgroup$ Dec 7 '19 at 17:55
  • $\begingroup$ However, I think you meant to write $\binom{95}{9}$ instead of $\binom{95}{1}$ in the last equation. I am not able to edit this, because edits must change a minimum of six characters and those would just be two. $\endgroup$ Dec 7 '19 at 17:56
  • $\begingroup$ Indeed, I did. Thank you for pointing out the error. $\endgroup$ Dec 7 '19 at 18:01
0
$\begingroup$

Using conditional probability (imagine picking the 5 songs in order), you want this product:

P(s1 non-B)*P(s2 nonB|s1 nonB)*P(s3 nonB|s1,s2 nonB)*P(s4 nonB|s1,s2,s3 nonB)*P(s5 B|s1-s4 nonB) =

$(90/100)(89/99)(88/98)(87/97)(10/96) = 0.067878...$ as above.

$\endgroup$
1
  • $\begingroup$ Thank you, but this is a restatement of Devore's first method, which I had no doubts about. The question was about his second approach. $\endgroup$ Dec 8 '19 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.