Engineering Probability Class 12 and Exam 1 Answers - Thu 2019-02-21

Source

Name, RCSID:

WRF answers

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 six 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. You are trying to pass your driving test to get a driving license. You can take the test only once a month. For each time you try, you pass with probability 1/3. The random variable is the number of months until you pass for the first time.

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

      geometric

    2. (5 pts) What's the expected number of months until you pass?

      p=1/3, mean=1/p. So 3.

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

      sqrt(1-p)/p = 3 sqrt(2/3) = 2.4

  2. In this year 2019, which has 365 days, the set of outcomes is the set of days of the year, from Jan 1 to Dec 31. Event A is that the day is Monday. Event B is that the day is in Jan. Here is a calendar for Jan:

        January 2019
Su Mo Tu We Th Fr Sa
       1  2  3  4  5
 6  7  8  9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30 31

    1. (5 pts) What is the probability of B?

      31/365 = .0849

    2. (5 pts) Are A and B independent? Prove your answer (don't just state it).

      One year is 52 weeks plus one day. In 2019, the extra day is Tues.

      P(A) = 52/365=.1424

      P(A and B) = 4/365 = .0109. That does not equal P(A)P(B)=.0120

      no independent.

  3. You are scanning a B&W page and transmitting it over a noisy channel. A black bit is coded as 1 and a white bit as 0. Event A is that a random bit is black. P(A) = .01. Event B is that the bit is received as black. Sometimes the bit is changed during transmission. 1% of the white bits are changed to black. But 10% of the black bits are changed to white.

    1. (5 pts) What is the probability of B?

      P(B)=P(B|A)P(A)+P(B|A ' )P(A ' ) = .9*.01+.01*.99 = .0189

    2. (5 pts) What is P(A and B))?

      P(A and B) = P(B|A) P(A) = .9 .01 = .009

    3. (5 pts) What is P(A' and B')?

      P(A' and B') = P(B' | A' ) P(A') = .99 * .99 = .9801

    4. (5 pts) What is the probability that the bit arrived correct?

      P(A and B) + P(A' and B') = .9891

    5. (5 pts) What is P(A|B), the probability that 1 was transmitted, if you received 1?

      P(A|B) = P(A and B)/P(B) = .009/.0189 = .4761

    6. (5 pts) What is P(A' | B'), the probability that 0 was transmitted, if you received 0?

      P(A' | B') = P(A' and B') / P(B') = .9801 / .9811 = .9989

    7. (5 pts) Now you take that received page and transmit it a second time over the same noisy channel. Let event C be that you receive a black bit the 2nd time. Compute P(C).

      P(C|B) = P(B|A), P(C|B') = P(B|A')

      P(C) = P(C|B) P(B) + P(C|B') P(B') = .9 * .0189 + .01 * .9811 = .0268

      That is, with the noise, as you repeatedly retransmit, black bits become more likely.

  4. (5 pts) An LCD display has 1000 * 1000 pixels. A display is accepted if it has 15 or fewer faulty pixels. The probability that a pixel is faulty coming out of the production line is 1e-5. Find the proportion of displays that are accepted.

    This would be a Poisson distribution with the random variable being the number of bad pixels in a display. The parameter, a=1e6 * 1e-5 = 10. We want P(X<=15). That is

    \(\sum_{k=0}^{15} e^{-10} 10^k / k! = .9513\)

  5. Pretend that there are only 28 students in this class. This is about their birthdays. Assume that they are uniformly distributed over the year (although this is actually false.)

    1. (5 pts) What's the probability that no one has a birthday on Feb 29? (Year 2000 was a leap year.) As always a reasonable formula is ok.

      (365/366)^28 = .9244

    2. (5 pts) Now pretend that no one has a birthday on Feb 29. What's the probability everyone has a different birthday?

      1st person is always ok.

      P(2nd person's birthday is different) = 364/365

      P(3rd is different from previous 2) = 363/365

      P(k-th student is different from 1st k-1) = (366-k)/365

      P(all different) = product = 365!/338!/365^28 = .3455

    3. (5 pts) Assume that RPI has 7000 students, and their birthdays are uniformly distributed and no one is on Feb 29. Use a reasonable approximate distribution to compute the probability that exactly 20 students' birthday is today, Feb 21.

      Use Poisson. The expected number of students with a birthday today is a=7000/365= 19.17.

      The probability of exactly 20 is \(e^{-19.17} 19.17^{20}/20!\) = .0873

  6. (5 pts) Able and Baker take turns tossing a coin until one gets Head. The winner is the person who made that toss. So, if Able gets a head on the first toss, he wins. If Able tosses Tail and then Baker tosses Head, Baker wins. And so on. What's the probability that Able eventually wins the game?

    P(Able wins) = P(Able wins on 1st toss) + P(wins on 3rd) + P(5th) + ...

    = 1/2 + 1/8 + 1/32 + ... = 2/3

End of exam 1, total 70 points.