1999 fall solutions

1999 fall solutions - Math 218 Final Exam Solved Fall 1999...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 218 Final Exam Solved Fall 1999 The solutions which we reproduce below are far more elaborate than what was expected on the final. After all, not many students have laser printers and Mathematica running on a laptop during the exam! Our purpose is to give as full an explanation as possible, so you understand the solution. Problem 1 . (15 points) A certain investor is interested in a group of 30 stocks. Among these stocks are 17 stocks traded at the New York Stock Exchange (NYSE) and 13 stocks traded at the American Stock Exchange (Amex): Among the NYSE stocks, the prices of 12 stocks are up since the previous trading day and the prices of 5 stocks are down. Among the Amex stocks, the prices of 7 stocks are up since the previous trading day and the prices of 6 stocks are down. When appropriate, you may leave your answer in terms of factorials. (a) Suppose one stock is selected at random. Let A represent the event that a stock traded on the NYSE is chosen, and let B represent the event that a stock whose price has risen is chosen. Find P ( A ) and P ( A | B ) . (b) Are A and B independent? Show your work. (c) The investor needs to read a report on each of the Amex stocks whose price is up since the previous trading day, and must choose the order in which to read them. How many ways are there for the investor to do this? Show your work. (d) How many ways can the investor choose a group of 5 different stocks to buy from the group of 13 Amex stocks? (e) Suppose that the investor wants to invest in some of the 13 Amex stocks. How many ways can the investor choose a group of 3 Amex stocks that are up and a group of 2 Amex stocks that are down? Solution. (a) There are a total of 30 stocks; the event A represents choosing one of the 17 NYSE stocks, so P ( A ) = 17 / 30 = 0 . 5667, the number of successes over the number of possibilities. By definition, P ( A | B ) = P ( A and B )/ P ( B ) , where B is the event “the price has risen”. There are 19 stocks which are up, of which 12 are NYSE stocks; thus P ( A and B ) = 12 / 30, and P ( B ) = 19 / 30. Therefore P ( A | B ) = 12 / 19 = 0 . 6316. Of course, this has also the common-sensical solution: probabilities given B are over a new uni- verse of 19 simple events (the stocks that went up), and 12 of them were from the NYSE, thus the probability of choosing an NYSE stock from the 19 which went up is 12 / 19. (b) For A and B to be independent, we must have P ( A and B ) = P ( A ) P ( B ) (because that’s the definition of independent events). So we are asking whether 12 19 = 17 30 × 19 30 , and this is certainly false. (For one thing, the numerator on the right side isn’t divisible by 3, and the numerator on the left is. Or pull out your calculator and check.) (c) 7 of the Amex stocks are up, and you’re asked how many ways the investor can order these 7 stocks. The answer is 7! = 1 × 2 × 3 × 4 × 5 × 6 × 7 = 5040 Perhaps he should quit daydreaming and read them alphabetically ;-) (d) This is ( 13 5 ) (which is even pronounced “13 choose 5”), or µ 13 5 = 13 · 12 · 11 · 10 · 9 1 · 2 · 3 · 4 · 5 = 1287 ways. (You can also use µ 13 5 = 13!
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 8

1999 fall solutions - Math 218 Final Exam Solved Fall 1999...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online