 The Clarinet BBoard
|
Author: Jack Kissinger
Date: 2009-04-09 22:57
This is apparently floating around the St. Louis Symphony (must not have any Rouse on the program this week). Don't know where they got it. Enjoy:
Math/Logic Quiz
1. Wilson is tired of paying for clarinet reeds. If he adopts a policy of playing only on rejected reeds from his colleagues will he be able to retire on the money he has saved if he invests it in mutual bonds, yielding 8.7%, before he is fired from his job? If not, calculate the probability of his ever working in a professional symphony orchestra again!
2. Jethro has been playing the double bass in a symphony orchestra for 12 years, three months and seven days. Each day, his inclination to practice decreases by the formula: (total days in the orchestra) x 0.0076. Assuming he stopped practicing altogether four years, six months and three days ago, how long will it be before he is completely unable to play the double bass?
3. Wilma plays in the second violin section, but specializes in making disparaging remarks about conductors and other musicians. The probability of her making a negative comment about any given musician is 4 chances in 7, and for conductors is 16 chances out of 17. If there are 103 musicians in the orchestra and the orchestra sees 26 different conductors each year, how many negative remarks does Wilma make in a two-year period? How does this change if five of the musicians are also conductors? What if six of the conductors are also musicians?
4. Horace is the General Manager of an important symphony orchestra. He tries to hear at least four concerts a year. Assuming that at each concert the orchestra plays a minimum of three pieces per concert, what are the chances that Horace can avoid hearing a single work by Mozart, Beethoven or Brahms in the next ten years?
5. Betty plays in the viola section. Despite her best efforts she is unable to play with the rest of the orchestra and, on average, plays 0.3528 seconds behind the rest of the viola section, which is already 0.16485 seconds behind the rest of the orchestra. If the orchestra is moving into a new concert hall with a reverberation time of 2.7 seconds, will she be able to continue playing this way undetected?
6. Ralph loves to drink coffee. Each week he drinks three more cups of coffee than Harold, who drinks exactly one third the amount that the entire brass section consumes in beer. How much longer is Ralph going to live?
7. Rosemary is unable to play in keys with more than three sharps or flats without making an inordinate number of mistakes. Because her colleagues in the cello section are also struggling in these passages she has so far been able to escape detection. What is the total number of hours they would all have to practice to play the complete works of Richard Strauss?
|
|
Reply To Message
|
|
Author: chris moffatt
Date: 2009-04-10 02:20
well Jack I cudgelled the old Math/IT major brain (what's left of it) and here are the answers:
1.Wilson: NO. Zero. He gets fired becomes an investment banker gets a federal bailout package and retires with a 20 billion dollar retention bonus.
2. Jethro hasn't played a note for 3years 4 months and 17 days. The sympathetic vibrations from the other three basses transmitted through the floor have made his bass resonate enough that no-one has noticed. If they do no problem, Wilson has offered him a job at the bank
3.Wilma:8757.78151260507 - this is an approximation
4.Horace; Zero. The first concert he attends will feature Beethoven's Pastoral, Eine kleine Nachtmusik and the Requiem by you know who.
5.Betty; she will start to use a mute, be congratulated on her delicate touch and be promoted to leader of the viola section. Then all members will start to use mutes - end of problem. no one will notice.
6. Ralph: Harvard Medical School and JAMA publish research that coffee is good for you. Ralph lives to be 103 but the brass section members are all fired for bringing beer onto the stage. They are replaced by the Ukelele orchestra of the UK. Concert attendance increases by 783% in 2 weeks.
7. Rosemary (of the 3 sharps & flats): 147 years,7 months and 28 days assuming they practice 6 hours/day 6 days/week, 51 weeks/year AND no hitherto unknown manuscripts by R Strauss are discovered.
|
|
Reply To Message
|
|
Author: EEBaum
Date: 2009-04-10 06:26
The mathematician in me is disappointed in the lack of sufficient information to solve many of these problems. However, I then realized that at least some of them are solvable with either abstract methodologies or by defining possible ranges for unspecified numbers.
My analysis...
1. The term "mutual bonds" does not appear in Wikipedia, and therefore does not exist. If he invested in mutual funds, Wilson has just lost the $328,416.23 due to the housing crash, and will not ever retire. He will be fired shortly, largely because of the drastic effects that his near-suicidal depression has had on his playing.
2. Initially, the depreciation formula seems irrelevant to the problem. However, it is in fact the most relevant part indeed. One would initially assume that someone who plays in a symphony orchestra would retain enough skill to at least be able to play the instrument, without practicing at all, just with the time spent playing during rehearsals and concerts. So, one might say, he will never become unable to play, barring death or physical injury, as long as he keeps with the orchestra.
According to the numbers provided, though, if Jethro's inclination to practice was initially 100%, or 1, he should have lost said inclination entirely in 16 days. In fact, after day 131, assuming he somehow had rejuvenated to 100%, his inclination would reach 0 by the very next day. The only way to reconcile this discrepancy is to assume that the orchestra meets very seldom indeed. Seeing has he stopped practicing altogether after almost 8 years with the orchestra, it can only be assumed that Jethro is "in the orchestra" for only two days out of the year. Without practicing outside of that, it is highly unlikely that Jethro can play the instrument at all, having only touched it a total of 8 times in the past four years.
3. This problem is particularly misleading, as it seems to suggest a correlation between the frequency of Wilma's remarks and the makeup of the group's personnel. The only relevant factor of this problem is the one that is suspiciously absent: The frequency and duration of rehearsals. A constant can be determined for the frequency of a second violinist's disparaging remarks (in RPH, i.e. Remarks Per Hour), with tables of such contents available both as approximated sectional averages and exact values by seat placement, but this data is useless without an indication of total rehearsal time. Unsolvable, not enough information.
4. A tricky problem, leading the solver to do a whole lot of unnecessary calculations. At the lower end, any major "important" orchestra has at LEAST a 1/4 chance of having a piece by Mozart, Beethoven, or Brahms on each concert, regardless of how many pieces are programmed. Over 40 concerts, this alone provides a (1/4) + (3/4 * 1/4) + (3/4 * 3/4 * 1/4) etc. probability of seeing Mozart, Beethoven, or Brahms. Imagine flipping a pair of coins 40 times and never having tails show up on both coins. One could even argue for the laughable possibility of a lower probability of hearing these composers, at which point the equation would be altered to a lower probability. Regardless, however, these calculations are rendered moot by the simple fact that all three composers will have a multiple-of-ten birthday celebrated within the next ten years, or any ten years for that matter (the "ten years" figure being the key to this problem). The probability of Horace going to four concerts of a major important orchestra during one of THOSE years without seeing Mozart, Beethoven, or Brahms is 0.
5. The reverberation of the hall is insignificant, provided that Wilma is a section violist (not principal). Reverberation is irrelevant. If nobody has noticed yet, nobody will continue to notice. If, however, the new hall employs a Jumbo-tron, one may worry that an obsessive-compulsive violist in the audience may cause some troubles indeed for dear Betty. However, this can very easily be dismissed as a discrepancy due to the delay of sound propagation. Betty's incompetence will remain undetected.
6. Provided that the brass section is not averse to alcohol consumption for reasons religious or otherwise, Ralph is dead already, as is Harold. At a very conservative estimate, two kegs plus three cups (as well as just two kegs) per week of coffee is a fatal dose of caffeine. This, of course, assumes that Ralph does not drink decaf, in which case he would simply never leave the restroom. In either case, he will no longer be with the orchestra.
7. I apologize that I am unable to solve this problem. All relevant data has been provided, but until I can locate my Calculus textbook and refresh my memory on the mathematical interactions of different kinds of infinity, I'll have to leave it unsolved. While not exact, Chris has provided a reasonable approximation.
-Alex
www.mostlydifferent.com
|
|
Reply To Message
|
|
Author: Ralph Katz
Date: 2009-04-10 13:12
The Institute for Fairness in Jokes has rated your quiz:
A) gender: 4 males, 3 females - Rating: excellent
B) specialty distribution: 1 clarinet, 1 double bass, 1 violin (2nd), 1 manager, 1 viola, 1 cello, 1 unknown - Rating: fair - 4 strings, no brass, no percussion.
C) disadvantage: 1 rhythm-limited, 1 key-limited, 1 poor attitude - Rating: poor - disparaging remarks about key-challenged player, disparaging portrayal of rhythm-challenged player. This sample represents a skewed percentage with regard to the typical professional orchestra.
D) humor quality: 1 snigger, 2 guffaws, 1 supressed laugh, 3 LOL's - Rating: Excellent.
Post Edited (2009-04-10 13:12)
|
|
Reply To Message
|
|
Author: Jack Kissinger
Date: 2009-04-10 18:49
A few comments to this point: (Be warned, Brenda, this will probably add cramps to your headache. )
1. I can't take credit for this quiz. One of the members of the St. Louis Symphony shared it with the president of a community orchestra I play with. He sent it to us.
2. Re: (1):
While the (verifiable) presence in Wikipedia of the term "Mutual Bonds" might constitute conclusive evidence that such bonds do exist, any mathematician/logician worth his/her salt could tell you that the absence of the term from Wikipedia is no guarantee that it does not exist unless Wikipedia is collectively exhaustive, which it is not. Furthermore even if there is no such type of bond as a Mutual Bond, as a quick check with Google will verify, there are a number of companies with the word "Mutual" in their title that could have issued bonds. These might also be identified as "Mutual" bonds where Mutual refers to the name of the issuer rather than the type of bond. This does not necessarily change the answer.
Assuming he goes through 50 boxes of reeds a year and his reeds cost $25 per box it would take about 52 years for him to amass the $1,250,000 Yahoo recently said he would need to retire (1,250,000 = (1250)(FV,.087,52). Unless he's Stanley Drucker or has money from another source, it ain't gonna happen.
3. Re: (4)
It seems to me that (4) might have an answer but we would need a lot of additional information. First of all, does the manager select concerts at random or deliberately. If the latter, if he consciously tries to avoid these three composers, then he probably can do so for most if not all major orchestras. On the other hand, if he consciously tries to attend concerts that include at least one work by one of the three, then he probably can hear them in every concert he attends in any 10-year period.
Assuming the manager's choice of concert is totally random, if we are willing to assume that programming is also totally random and we know the individual probabilities that each time the music director selects a piece, it will be by one of the three composers (MBB), then we can compute the probability that any 4 x 3 x 12 = 120 combinations of works (the number of works our manager will hear over 10 years) will include at least one work by MBB. Alex states that (at the low end) the odds any concert by a major orchestra will include a work by one of our three are 1/4. If the 2009-10 season of the SLSO is representative (and I use them because they provided this quiz), he's right. The odds for SLSO next year are actually somewhere between 36% and 38% depending on whether you calculate on a concert-by-concert or program-by-program basis (since 4 programs are only performed once, 13 are performed twice and 11 are performed 3 times and, BTW, the probablility that one of our icons will included on a program is a direct function of the number of times the program will performed).
Assume then that SLSO's programming is totally random and that there is a 36% probability that a work by MMB will be performed. The probability that at least one work by MBB will be performed is equal to 1 - the probability that no works by MBB will be performed or .64. If we assume that all concerts include three works (which, in reality, they do not), the probability that a work by someone other than MBB will be chosen at random (P[MBB']) is given by the formula:
.64 = P[MBB']^3
So the probability that a work by someone other than MBB will be chosen at random is about .90 and the probability that a work by MBB will be chosen is .10. If the concerts are randomly programmed, the probability that the 120 works our manager hears will include at least one work by MBB is: (1 - (.90)^120) or about .9999968. In other words, he's doomed.
(Assuming that pieces are selected, at least for a given year, without replacement, the theoretically correct approach, IMO makes the problem intractable.)
Of course orchestra program directors do not program randomly. Another way of attacking this problem would be to assume that the manager chooses his concerts each year randomly or, alternatively chooses the weeks he will attend randomly and the SLSO's programming is representative. Using SLSO's upcoming season, even trying to work the problem on a concert by concert basis becomes a computational nightmare, however, because, as I mentioned above, the orchestra does not perform each concert program the same number of times. If, on the other hand we are willing to assume that our manager is with the SLSO and chooses his weeks randomly (with equal probability), we can calculate the probability that he will not encounter MBB next year. To the extent that 2009-10 is representative of our mythical 10-year period (which, as Alex points out is a stretch), we can then extend next year's results to the 10-year period.
According to it's schedule for next year, the SLSO will have a 25-week subscription season. In 16 of those weeks, no work by MBB is performed. In 8 of the weeks, every concert performed includes at least one MBB work. Just to complicate matters, in the remaining week, one of the three concerts performed includes a work by MBB but the other two do not.
The probability that our manager will encounter MBB at least once next year = 1 minus the probability that he will not encounter MBB at all. The probability that he will not encounter MBB equals the sum of the probabilities that:
(1) he does not encounter MBB given that he does not choose the odd week x probability that he does not choose the odd week, and
(2) he does not encounter MBB given that he does choose the odd week x probability that he does choose the odd week.
The former probability is given by:
(17/25)(16/24)(15/23)(14/22)(21/25) = (about) 15.8%
The latter probability is given by:
(16/25)(15/24)(14/23)(13/22)(4/25) = (about) 1.15%
Thus the probability that he does not encounter MBB during the year is about 17%.
Assuming the schedule is about the same each year, and that each year's schedule is independent of the other years (perhaps not true but necessary to make the problem reasonably tractable) the probability that he will not encounter MBB in a 10-year period would be .17^10 = 2 E08. So the probability that he will encounter MBB is .999999998. And he's still doomed.
Chris' solution, while perhaps somewhat cynical, is far more elegant.
Best regards,
jnk
|
|
Reply To Message
|
|
Author: Don Berger
Date: 2009-04-10 20:51
Interesting, BUT I'm with you, Brenda and Ralph, I just keep hoping that Some Day my brain will accept consideration of differential equations [simple ones] again. My logic is now limited to AND, OR, and perhaps their N---s. Don
Thanx, Mark, Don
|
|
Reply To Message
|
|
The Clarinet Pages
|
 |
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
.jpg){kind=small}
