Engineering Probability 2018 Exam 1 Solution - Mon 2018-02-26

Source

Name, RCSID: WRF solutions

Note: Full points will be given for an expression with the numbers, w/o computing the answer.

Rules:

  1. You have 80 minutes.
  2. You may bring one 2-sided 8.5"x11" paper with notes.
  3. You may bring a calculator.
  4. You may not share material with each other during the exam.
  5. No collaboration or communication (except with the staff) is allowed.
  6. Check that your copy of this test has all seven pages.
  7. Do any 14 of the 17 questions or subquestions. Cross out the 3 that you don't do.
  8. When answering a question, don't just state your answer, prove it.

Questions:

  1. (5 pts) Ten people run a race for gold, silver, bronze. How many ways can the medals be won, w/o any ties?

    10*9*8 = 720.

  2. Two teams, the Albanians and the Bostonians, are playing a 7 game series. The first team to win 4 games wins the series, and no more games are played. In any game, the Albanians have a 60% chance of winning. The games are independent, and there are no ties.

    1. (5 pts) What's the probability that the series will run to 7 games?

      The first 6 games must have exactly 3 Albanian wins.

      \(p = {6 \choose 3} .6^3 .4^3 = .276\)

    2. (5 pts) What's the probability that the Albanians win the series?

      They might win in

      1. 4 games with prob \(.6^4=0.130\) or in
      2. 5 games with prob \({4 \choose 3} .6^4 .4=0.201\)
      3. 6 games with prob \({5 \choose 3} .6^4 .4^2=0.207\)
      4. 7 games with prob \({6 \choose 3} .6^4 .4^3=0.166\)

      The sum is 0.704 .

  3. (5 pts) Imagine two coins. Coin A has two heads. Coin B has the usual one head and one tail, and it is fair. You pick a coin at random (p=.5 to pick either coin) and toss it. It comes up heads.

    What is the probability that you picked coin A?

    P[A] = P[A'] = 1/2. P[H|A] = 1. P[H|A'] = 1/2.

    P[A & H] = P[A] P[H|A] = 1/2.

    P[A' & H] = P[A'] P[H|A'] = 1/4.

    P[H] = P[A&H]+P[A'H] = 3/4.

    P[A|H] = P[A&H]/P[H] = (1/2)/(3/4) = 2/3.

  4. (5 pts) Consider S={1,2,3,...22}. Is the set of even numbers independent of the set of multiples of 4?

    There are 11 even numbers in S, 5 multiples of 4, and 5 both.

    P[even] = 11/22 =1/2. P[mult] = 5/22. P[even]*P[mult] = 5/44.

    P[even and mult] = 5/22 not = 5/44.

    They are not independent.

  5. You are sitting an online multiple choice exam in thraumaturgy, about which you know nothing. Each question has 5 possible answers. You answer each question randomly. The exam ends when you get your first correct answer.

    1. (5 pts) What's the relevant probability distribution?

      Geometric.

    2. (5 pts) What's the expected number of questions you will need to answer?

      p=1/5

      E = 1/p = 5.

    3. (5 pts) What's the standard deviation?

      STD = sqrt(1-p)/p = 4.47.

  6. You are designing a check bit system for transmitting 8-bit bytes over a noisy channel. Since it's really noisy, you append two check bits to each byte, transmitting ten bits in total for each byte. Each bit, independently, can be wrong with probability \(10^{-6}\).

    You're using a Reed-Solomon error correction scheme, which we teach in another class. If there is only one bad bit in the ten transmitted bits, it will correct the byte. If there are two bad bits, it will report the error, but can't correct it. Three or more bad bits are unlikely enough that we assume they never occur.

    1. (5 pts) What's the probability that the receiver can deduce the correct byte?

      That would be the probability that 9 or 10 bits are ok.

      Let p = prob a given bit is bad. \(p=10^{-6}\)

      Let q=1-p = 0.999999.

      P[all 10 bits ok] = \(q^{10}\)

      P[exactly 9 ok] = \({10 \choose 1} p q^9\)

      P[9 or 10 ok] = \(q^{10}+{10 \choose 1} p q^9 = q^9 (q+10p)= .999999999955000\)

    2. (5 pts) What's the probability that the receiver will receive a byte that it knows is bad, but can't correct?

      prob of exactly 2 errors of the 10 bits.

      \({10 \choose 2} p^2 q^8 = 45 \cdot 10^{-12} \cdot 0.999999^8 = 4.5\cdot10^{-11}\)

      Note: These probabilities are small, but if you are transmitting millions of bytes, then they're significant. If you didn't add extra bits, the probability of an 8-bit byte having a bad bit is \(1-q^8=8\cdot10^{-6}\). The error correction reduced the probability of a bad byte by a factor of 50,000, at a cost of 25% more transmission and some computation. More, if the 8-bit byte is bad, you don't know it. However, if the 10-bit byte has 2 bad bits, you do know it. Those are the advantages of error correcting codes.

  7. Pretend that we divide the 86 field into a grid of 100 by 100 squares. 5000 students toss 10 paper airplanes each off the JEC roof. Each paper airplane has an independent and uniform probability of hitting each square. Each airplane falls into exactly one square.

    1. (5 pts) What's the mean number of airplanes to hit a particular square?

      50000 airplanes, 10000 squares. mean = 5.

    2. (5 pts) What's the exact probability that a particular square gets zero airplanes? It's ok to give an expression; you don't need to evaluate it.

      Let p = prob this square gets a particular airplane = 1/10000 .

      Let q=1-p.

      Prob this particular square gets no airplane = \(q^{50000}=.0067362626\) .

    3. (5 pts) What's a faster very good approximate formula? An expression is ok.

      Poisson is appropriate here. Call the mean a. a=5.

      \(P[0] = a^0 e^{-a}/ 0! = e^{-5}\approx .007\)

    4. (5 pts) What's the very good approximate standard deviation for the number of airplanes to hit a particular square? An expression is ok.

      Variance equals mean, so std = \(\sqrt{5}\)

  8. Chip quality control:

    1. Each chip is either good or bad.
    2. P[good]= 0.9.
    3. If the chip is good: P[still alive at t] = \(2^{-t}\)
    4. If the chip is bad: P[still alive at t] = \(3^{-t}\)

    Questions:

    1. (5 pts) What's the probability that a random chip is still alive at t=2? Give an expression and evaluate it to give a number.

      Let G = good. Let A = event that chip is alive at 2.

      P[A|G] = 1/4. P[A|G'] = 1/9. P[A&G] = .25 * .9 = .225.

      P[A&G'] = 1/9 * .1 = .01111.

      P[A] = P[A&G] + P[A&G'] = .236 .

    2. (5 pts) If a random chip is still alive at t=2, what's the probability that it's a good chip?

      Use Bayes.

      P[G|A] = P[A&G]/P[A] = .225/.236 = .95

    3. (5 pts) If a random chip is still alive at t=2, what's the probability that it will still be alive at t=3?

      Let B = event that chip is alive at 3.

      B implies A, so the event (B&A) is the event B.

      P[B|A] = P[B & A] / P[A] = P[B] / P[A]

      P[B|G] = 1/8. P[B|G'] = 1/27.

      P[B&G] = P[B|G] P[G] = 1/8 * .9 = .1125

      P[B&G'] = P[B|G'] P[G'] = 1/27 * .1 = 0.0037.

      P[B] = P[B&G] + P[B&G'] = 0.1125 + 0.0037 = 0.1162 .

      P[B|A] = P[B] / P[A] = 0.1162 / .236 = 0.492.

End of exam 1, total 70 points.