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General Betting
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02 Apr 10
11:36
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Date Joined:
20 Jan 05
Maths wiz here in Japan set us some brainteasers yesterday. Have to say apart from the first 2 I didn't do very well. A maths degree would be useful. Post up your answers and I'll put his answers up tomorrow...have fun!
1) If a baseball squad has 8 pitchers, 12 infielders, and 9 outfielders, how many ways can an MVP award be given to one player of each category?
2) When drawing four cards from a pack, what is the probability of getting one from each suit?
3) What is the probability of a randomly selected positive integer giving a number ending in 1, if raised to the fourth power?
4) Suppose n people sit down, randomly and independently, in an auditorium with n + k seats. Assuming m seats are specified in advance, what is the probability that they will be occupied? (m n).
5) In a batch of 400 tickets the scanner makes 8 errors. Assuming we pull 3 tickets randomly what is the probability that none of them will me among those not scanned? (And how big an error would we make if we approximated this probability with the binomial probability "zero successes in three tirals" when p = 8/400 = 0.02?)
1) If a baseball squad has 8 pitchers, 12 infielders, and 9 outfielders, how many ways can an MVP award be given to one player of each category?
2) When drawing four cards from a pack, what is the probability of getting one from each suit?
3) What is the probability of a randomly selected positive integer giving a number ending in 1, if raised to the fourth power?
4) Suppose n people sit down, randomly and independently, in an auditorium with n + k seats. Assuming m seats are specified in advance, what is the probability that they will be occupied? (m n).
5) In a batch of 400 tickets the scanner makes 8 errors. Assuming we pull 3 tickets randomly what is the probability that none of them will me among those not scanned? (And how big an error would we make if we approximated this probability with the binomial probability "zero successes in three tirals" when p = 8/400 = 0.02?)
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xxlandofheadachesxx
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1 8x9x12= whatever is is
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I think the answer to (3) is 40%, but i'm unable to explain why!
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39/51 x 26/50 x 13/49 = whatever it is = p getting one from each suit... first card chosen is irreletive
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3 .. how many number raised by the fourth power end in 1 first.. will only be odd numbers..
therefore 1 does.. 3 does 5 no 7 does 9 does therefore 80 % |
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^agree with the above
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392/400 x 391/399 x 390 x 398 = prob not drawing bad ticket
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If you're looking through the accounts of a random company and you're looking at a random expense, what is the chance that the first digit is a 1.
Note: I know the answer to this. It's not 10% Note 2: I won't be able to explain it, as the maths is in a field I suck at horribly and I only have a vague grasp of why it works. |
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3 .. how many number raised by the fourth power end in 1 first.. will only be odd numbers..
therefore 1 does.. 3 does 5 no 7 does 9 does therefore 80 % Surely 40% because only 4 of the 10 numbers between 0 and 9 do |
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1 chance in 11 lori?
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Question 1 I don't understand this. How can there be more that one "most valuable player" assuming that is what MVP means.
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no wait that can't be right
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loll... yep sorry lay low..ur right
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an expense is usually a number of something multiplied by a cost... it is only certain numbers that when times together result in a 1...
1 x 1 .... 9 x 9 ... 3x 7 etc.... certain numbers for example 7 are less common since there are not as many common variables that when times together result in 7... therefore i would imagine 1 being the last number on and expenses sheet would be nearer 20 % than 10 % thats my logic anyway |
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if you did a grid with 1 x 1 1 x 2 1x 3 down one colum then 2 x1 2 x 2 2 x 3 etc down next etc etc all the way to 1x9 and 2 x 9 and did all coloums then this would explain it... btw i imagine 0 is the top answer with 25 % roughly
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Lori, I've seen that mentioned, it's Benford's law http://en.wikipedia.org/wiki/Benford%27s_law
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How about the first number Sherrif?
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That's the one Fred ;)
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*Sheriff
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lol... note to self... alwasy read the question asked not the question you woiuld like to be asked...
ill have think about it... |
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always remember benfords law when fiddling expenses ;)
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1) If a baseball squad has 8 pitchers, 12 infielders, and 9 outfielders, how many ways can an MVP award be given to one player of each category?
This is simply 8*12*9 = 864 different ways 2) When drawing four cards from a pack, what is the probability of getting one from each suit? The chance of getting a new suit is 100% on the first draw, then it is 39/51, then 26/50, and finally 13/49, which is 13/17*13/25*13/49 = 2197/20,825 or 10.55%. 3) What is the probability of a randomly selected positive integer giving a number ending in 1, if raised to the fourth power? It is .4, or 40% as any numbers ending in 1, 3, 7, or 9 when raised to the fourth power give a number ending in 1. 4) Suppose n people sit down, randomly and independently, in an auditorium with n + k seats. Assuming m seats are specified in advance, what is the probability that they will be occupied? (m n). I find this easier to explain with an example. Let's make n = 7, k = 6, and m = 3. There are (13*12*11*10*9*8)/6! ways for the n to sit in the auditorium. If m seats are all occupied then the remaining seats are 10 (13-3) choose 4 (7-3) , or (10*9*8*7)/4!. The first number is 1716, the second 420, so it is 105/429 a seat is occupied. I'm afraid if I tried to convert that back into algebra I would wind up with alphabet soup, but swap in any values for k, m, n and follow the steps to the answer. 5) In a batch of 400 tickets the scanner makes 8 errors. Assuming we pull 3 tickets randomly what is the probability that none of them will be among those not scanned? (And how big an error would we make if we approximated this probability with the binomial probability "zero successes in three trials" when p = 8/400 = 0.02?) There are 392 chances in 400 on the first pull (49/50), 391/399 on the second, and 390/398 on the third, so I get .941047. The answer to the second part involves, I think, a table lookup. If I read the tables correctly the approximate probability approaches 100%. Probably I read the tables wrong, but otherwise the error is nearly 6%. |