ECSE-2500 Engineering Probability, RPI, Spring 2011 Home Page

ECSE-2500 Engineering Probability, Exam 1, March 1, 2011

NAME: _____ answers

EMAIL:___________________________ RIN:______________

Answer every question. There are four pages with four questions, total 68 points.

You may write free for any 6 points.

We implemented this by adding 6 points to everyone and capping the total at 68.

This exam is closed book but a calculator and one 2-sided letter-paper-size note sheet are allowed. You may not use other computers or communication devices, or share material with other students.

In the following questions, the life of a light is an integer number of weeks, not a real number. I.e., these distributions are discrete not continuous.
You have a stockroom of 2000 hi quality and 1000 lo quality light bulbs. Unfortunately they are unlabeled. You take one light, but don't know which it is. H is the event that it's a hi quality light.
The probability that a hi quality light dies after k whole weeks (i.e., in the k+1-st week) is geometric. The probability that it dies after 0 whole weeks is 0.1.
For a lo quality light, that probability is 0.2.
1. [2 pts] What's the mean number of weeks that a hi quality light will work?
  Prob of dying after exactly k whole weeks: {$ p=.1, p_k = p (1-p)^k, k\ge0 $}
  {$ \mu_{hi} = (1-p)/p = 9. $}
  Reasonably close, i.e., off-by-1, errors are allowed.
2. [2] What's the probability that the light that you took was hi quality?
  2000/(2000+1000) = 2/3
3. [4] What's the mean life of the unknown light that you took?
  {$ \mu_{lo} = 4, \mu = \frac{2}{3} \mu_{hi} + \frac{1}{3} \mu_{lo} = \frac{22}{3} $}
4. [6] What's the probability that your unknown light dies before 2 weeks?
  Hi-q: {$ p_k=.9^{k-1}.1,\ \ \ \ p_0+p+1=.1+.09=.19 $}
  Lo-q: {$ p_k=.8^{k-1}.2, \ \ \ \ p_0+p+1=.2+.16=.36 $}
  Combine: {$ p= \frac{2}{3} .19 + \frac{1}{3} .36 = \frac{74}{300} $}
5. [4] What's the probability that the hi quality light dies before 2 weeks?
  Computed above: .19
6. [4] With a 2/3 probability, what's the range for its life? I.e., give the mean-std and mean+std.
  Reasonably close, i.e., off-by-1, errors are allowed.
  {$ \mu = 9, \sigma = \frac{1-p}{p^2} = \sqrt{90} = 9.5 $}
  The range is approx {$ \mu\pm\sigma = [0, 18.5] $}
  Reasonably close, i.e., off-by-1, errors are allowed.
7. [6] The unknown light that you took died after zero weeks. What's the probability that it was a hi quality light?
  Use Bayes. Let H be the event that it was a hi-q light. D is the event that it died after 0 weeks.
  P[H|D] = P[H ∩ D]/P[D] = 1/2
  P[D] = P[D|H]P[H] + P[D|H']P[H'] = .1x2/3 + .2x1/3 = 2/15
8. [6] The longer your unknown light lasts, the more likely that it was a hi quality light. There is some number n, that if the light dies after n weeks, then there was a 99% probability that the light was hi quality. How would you determine n. Don't actually compute n, just tell us in great detail how to.
  H = event that this was a hi-q light.
  D = event that light died after n weeks. Either exactly n, or n or more, are ok. I'll give the answer for n or more, since it's harder. Exactly n should be obvious.
  P[D|H] = {$ \sum_{k=n}^\infty .9^{k} .1 $}
  P[D|H'] = {$ \sum_{k=n}^\infty .8^{k} .2 $}
  P[H] = 2/3, P[H'] = 1/3
  P[D] = P[D|H]P[H] + P[D|H']P[H']
  P[D ∩ H] = P[D|H]P[H]
  P[H|D] = P[D ∩ H] / P[D]
  Plot P[H|D] for various n and see where it becomes 0.99.
9. [2] Let's start again, with the 2000 hi quality and 1000 lo quality lights. You select 4 lights. What's the probability that exactly 2 of them are hi quality? It's ok to use reasonable approximations.
  {$ {4 \choose 2} \left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^2 = \frac{64}{81} = \frac{8}{27} $}
You're trying to read my handwriting of a string of digits, but are having problems. (Perhaps you need new glasses.) Each digit is either 4 or 9, which can get confused.
If I write a 4, you see a 4 80% of the time, and a 9 20%.
If I write a 9 you see a 9 60% of the time and a 4 40%.
I'm writing 4s and 9s with equal probability.
1. [2] I write the string 499. What's the probability that you read 499?
  The 3 digits are independent. .8x.6x.6 = .288
2. [4] I write 3 digits. You think I wrote 499. What's the probability that I really wrote 499?
  Define W4 to be the event that I wrote a 4. S4 is the event that you saw a 4. Ditto for W9 and S9.
  P[W4] = P[W9] = .5
  P[S4|W4]=.8, P[S4 ∩ W4]=.4, P[S4|W9]=.4, P[S4 ∩ W4]=.2, P[S4]=.4+.2=.6, P[W4|S4]=.4/.6=2/3
  P[S9|W9]=.6, P[S9 ∩ W9]=.3, P[S9|W4]=.2, P[S9 ∩ W4]=.1, P[S9]=.3+.1=.4, P[W9|S9]=.3/.4=3/4.
  Answer = 2/3 x 3/4 x 3/4 = 9/24
You toss a fair 12-sided die once and look at the top face.
Event A occurs when that number is even.
P[A] = 1/2
Event B occurs when it is a multiple of 3.
P[B] = P[3,6,9,12] = 1/3
Event C occurs when it is 2, 4, 6, 7, 9, or 11.
P[C] = 1/2
1. [2] Are A and B independent? In each case here, justify your answer.
  P[A ∩ B] = P[6,12] = 1/6 Independent since = 1/2 x 1/3.
2. [2] A and C?
  P[A ∩ C] = P[2,4,6] = 1/4 = 1/2 x 1/2. Independent.
3. [2] B and C?
  P[B ∩ C] = P[6,9] = 1/6 = 1/3 x 1/2. Independent.
4. [4] A, B, and C?
  P[A ∩ B ∩ C] = P[9] = 1/12 = 1/2 x 1/2 x 1/3. Independent.
5. [2] Compute {$ P[A\cup B\cup C] $} by counting outcomes.
  {$ P[A\cup B\cup C] = \{2,3,4,6,7,8,9,10,11,12\} = 10/12 = 5/6$}
6. [4] Compute {$ P[A\cup B\cup C] $} from P[A] etc, {$ P[A\cap B] $} etc and {$ P[A\cap B\cap C] $}.
  {$ P[A\cup B\cup C] = P[A] + P[B] + P[C] - P[A\cap B] - P[A\cap C] $} {$ - P[B\cap C] + P[A\cap B\cap C] $}
  = 1/2 + 1/3 + 1/2 = 1/4 = 1/6 - 1/6 + 1/12 = 10/12 = 5/6.
You buy a can of uranium ore from Amazon for $39.95 plus $9.49 shipping. (This is a real product.) Its activity is 200 counts per minute. That means that a Geiger counter would click, on average 200 times per minute. However, you being busy count counts for only 15 seconds.
1. [2] What's the relevant probability distribution?
  Poisson. A large number of unlikely independent events.
2. [2] What's the value of the relevant parameter?
  {$\alpha$} , the mean number of decays per time unit. The time unit is 15 seconds here. {$\alpha = 200/4 = 50 $} .
3. [2] What are the mean and standard deviation of your count?
  {$\mu=\alpha=50,\ \sigma=\sqrt{\alpha}=7 $} .
4. [4] What's the probability that you will count less than 40 counts (in 15 seconds). Write this as a summation. You don't need to give a number.
  {$$ \sum_{k=0}^{39} \frac{50^k}{k!}e^{-50} $$}

(end of exam)