Engineering Probability Class 4 Thu 2019-01-24

W Randolph Franklin (WRF), RPI

2019-01-24 00:00

Source

Table of contents::

1 Handwritten notes from class
2 Chapter 2 ctd
3 Iclicker questions

1 Handwritten notes from class

here

E.g., 117.pdf

2 Chapter 2 ctd

Today: counting methods, Leon-Garcia section 2.3, page 41.
1. We have an urn with n balls.
2. Maybe the balls are all different, maybe not.
3. W/o looking, we take k balls out and look at them.
4. Maybe we put each ball back after looking at it, maybe not.
5. Suppose we took out one white and one green ball. Maybe we care about their order, so that's a different case from green then white, maybe not.
Applications:
1. How many ways can we divide a class of 12 students into 2 groups of 6?
2. How many ways can we pick 4 teams of 6 students from a class of 88 students (leaving 64 students behind)?
3. We pick 5 cards from a deck. What's the probability that they're all the same suit?
4. We're picking teams of 12 students, but now the order matters since they're playing baseball and that's the batting order.
5. We have 100 widgets; 10 are bad. We pick 5 widgets. What's the probability that none are bad? Exactly 1? More than 3?
6. In the approval voting scheme, you mark as many candidates as you please. The candidate with the most votes wins. How many different ways can you mark the ballot?
7. In preferential voting, you mark as many candidates as you please, but rank them 1,2,3,... How many different ways can you mark the ballot?
Leon-Garcia 2.3: Counting methods, pp 41-46.
1. finite sample space
2. each outcome equally probable
3. get some useful formulae
4. warmup: consider a multiple choice exam where 1st answer has 3 choices, 2nd answer has 5 choices and 3rd answer has 6 choices.
  1. Q: How many ways can a student answer the exam?
  2. A: 3x5x6
5. If there are k questions, and the i-th question has \(n_i\) answers then the number of possible combinations of answers is \(n_1n_2 .. n_k\)
2.3.1 Sampling WITH replacement and WITH ordering
1. Consider an urn with n different colored balls.
2. Repeat k times:
  1. Draw a ball.
  2. Write down its color.
  3. Put it back.
3. Number of distinct ordered k-tuples = \(n^k\)
Example 2.1.5. How many distinct ordered pairs for 2 balls from 5? 5*5.
iClicker. Suppose I want to eat one of the following 4 places, for tonight and again tomorrow, and don't care if I eat at the same place both times: Commons, Sage, Union, Knotty Pine. How many choices to I have where to eat?
1. 16
2. 12
3. 8
4. 4
5. something else
2.3.2 Sampling WITHOUT replacement and WITH ordering
1. Consider an urn with n different colored balls.
2. Repeat k times:
  1. Draw a ball.
  2. Write down its color.
  3. Don't put it back.
3. Number of distinct ordered k-tuples = n(n-1)(n-2)...(n-k+1)
iClicker. Suppose I want to visit two of the following four cities: Buffalo, Miami, Boston, New York. I don't want to visit one city twice, and the order matters. How many choices to I have how to visit?
1. 16
2. 12
3. 8
4. 4
5. something else
Example 2.1.6: Draw 2 balls from 5 w/o replacement.
1. 5 choices for 1st ball, 4 for 2nd. 20 outcomes.
2. Probability that 1st ball is larger?
3. List the 20 outcomes. 10 have 1st ball larger. P=1/2.
Example 2.1.7: Draw 3 balls from 5 with replacement. What's the probability they're all different?
1. P = \(\small \frac{\text{# cases where they're different}}{\text{# cases where I don't care}}\)
2. P = \(\small \frac{\text{# case w/o replacement}}{\text{# cases w replacement}}\)
3. P = \(\frac{5*4*3}{5*5*5}\)
2.3.3 Permutations of n distinct objects
1. Distinct means that you can tell the objects apart.
2. This is sampling w/o replacement for k=n
3. 1.2.3.4...n = n!
4. It grows fast. 1!=1, 2!=2, 3!=6, 4!=24, 5!=120, 6!=720, 7!=5040
5. Stirling approx:
  
  \begin{equation*} n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\left(1+\frac{1}{12n}+...\right) \end{equation*}
6. Therefore if you ignore the last term, the relative error is about 1/(12n).
Example 2.1.8. # permutations of 3 objects. 6!
Example 2.1.9. 12 airplane crashes last year. Assume independent, uniform, etc, etc. What's probability of exactly one in each month?
1. For each crash, let the outcome be its month.
2. Number of events for all 12 crashes = \(12^{12}\)
3. Number of events for 12 crashes in 12 different months = 12!
4. Probability = \(12!/(12^{12}) = 0.000054\)
5. Random does not mean evenly spaced.
2.3.4 Sampling w/o replacement and w/o ordering
1. We care what objects we pick but not the order
2. E.g., drawing a hand of cards.
3. term: Combinations of k objects selected from n. Binomial coefficient.
  
  \begin{equation*} C^n_k = {n \choose k} = \frac{n!}{k! (n-k)!} \end{equation*}
4. Permutations is when order matters.
Example 2.20. Select 2 from 5 w/o order. \(5\choose 2\)
Example 2.21 # permutations of k black and n-k white balls. This is choosing k from n.
Example 2.22. 10 of 50 items are bad. What's probability 5 of 10 selected randomly are bad?
1. # ways to have 10 bad items in 50 is \(50\choose 10\)
2. # ways to have exactly 5 bad is 3 ways to select 5 good from 40 times # ways to select 5 bad from 10 = \({40\choose5} {10\choose5}\)
3. Probability is ratio.
Multinomial coefficient: Partition n items into sets of size \(k_1, k_2, ... k_j, \sum k_i=n\)

\begin{equation*} \frac{n!}{k_1! k_2! ... k_j!} \end{equation*}
2.3.5. skip
More state lottery incompetence: Statistician Cracks Code For Lottery Tickets

Finding these stories is just too easy.

Reading: 2.4 Conditional probability, page 47-

3 Iclicker questions

Retransmitting a very noisy bit 2 times: The probability of each bit going bad is 0.4. What is probability of no error at all in the 2 transmissions?
1. 0.16
2. 0.4
3. 0.36
4. 0.48
5. 0.8
Flipping an unfair coin 2 times: The probability of each toss being heads is 0.4. What is probability of both tosses being tails?
1. 0.16
2. 0.4
3. 0.36
4. 0.48
5. 0.8
Flipping a fair coin until we get heads: How many times will it take until the probability of seeing a head is >=.8?
1. 1
2. 2
3. 3
4. 4
5. 5
This time, the coin is weighted so that p[H]=.6. How many times will it take until the probability of seeing a head is >=.8?
1. 1
2. 2
3. 3
4. 4
5. 5