breakingdown

breakingdown - Breaking Things Down 1. Introduction. By...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Page 1 Breaking Things Down 1. Introduction. By itself, the method of obstacles only works in a narrow range of cases. For example, suppose two tickets will be drawn, one at a time, at random from the box shown below: If you want to ± nd the chance the same letter shows on both tickets, the method of obstacles can handle this easily. The ± rst ticket can be any ticket but the second must have the same letter as the ± rst one: The answer is 1/3. Now suppose you want to ± nd the same chance with a slightly different box: You can’t use an obstacle course, because the chance the second let- ter is the same as the ± rst could be 1/4 or 1/2, depending on whether A A Z Z A A Z Z Z any ticket 4/4 same letter as ± rst 1/3
Background image of page 1

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

View Full DocumentRight Arrow Icon
Page 2 the letter on the f rst ticket is A or Z. Something else is required. But beFore reading on, you might try to guess which box gives a better chance oF getting the same letter twice: the Four ticket box or the f ve ticket box. Here is the method. You break down the possibility: same letter shows on both tickets into diFFerent ways it can happen: get A on both tickets or get Z on both tickets Next, you f nd the chances For these (using obstacles): get A on both tickets: 2/5 × 1/4 = 2/20 or get Z on both tickets: 3/5 × 2/4 = 6/20 And the last step is to add the chances oF the diFFerent ways: 2/20 + 6/20 = 8/20 = 2/5. That ends the method. It leads to a 2/5 chance the same letter will come up twice. This is bigger than the chance For the Four ticket box, which is 1/3. (You might have guessed the chance would go up by imagining an extreme case: that the box contained 2 As and 100 Zs, say. Then the chance would be almost certain both letters would be the same, because they would both be very likely to be Zs. So it looks like even adding a single Z to the box will put the chance up.) Example 1. Someone shuFfl es a deck oF cards and deals out, one at a time, f ve cards. ±ind the chance there is not more than one ace in the f ve cards. Do you think the chance will turn out to be closest to 50%, 75%, or 100%? Answer. ±irst, break down the possibility: not more than one ace as Follows. There won’t be more than one ace iF either: there are 0 aces in the f ve cards or there is 1 ace in the f ve cards There are two chances to f nd. The f rst one is straightForward, be-
Background image of page 2
Page 3 cause 0 aces in f ve cards can be written as an obstacle course: The chance there are 0 aces in the f ve cards is nearly 66%: 48/52 × 47/51 × 46/50 × 45/49 × 44/48 ≈ 0.6588 That does the f rst line oF the breakdown: not more than one ace in the f ve cards iF either: there are 0 aces in the f ve cards: 0.6588 or there is 1 ace in the f ve cards The second line takes most of the work. The possibility oF 1 ace (in the f
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 02/04/2011 for the course STAT 21 taught by Professor Anderes during the Fall '08 term at Berkeley.

Page1 / 22

breakingdown - Breaking Things Down 1. Introduction. By...

This preview shows document pages 1 - 4. Sign up to view the full document.

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