#Prob ECSE-2500-01 Engineering Probability, Spring 2018, Rensselaer Polytechnic Institute (Posts about exam)

Engineering Probability Exam 3 - Tues 2018-05-08

W Randolph Franklin (WRF), RPI — Mon, 07 May 2018 04:00:00 GMT

Name, RCSID:

.



.

Rules:

You have 80 minutes.
You may bring three 2-sided 8.5"x11" papers with notes.
You may bring a calculator.
You may not share material with each other during the exam.
No collaboration or communication (except with the staff) is allowed.
Check that your copy of this test has all nine pages.
Each part of a question is worth 5 points.
You may cross out three question parts, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

You toss two coins. Each comes up heads half of the time. However, for some funny reason, they both come up heads together, or both come up tails together. Intuitively, they not independent. This question is to prove that from the definition of independence.
```
.











.
```
This time, you toss three coins, A, B, and C. These are the probabilities:

P[TTT] = P[THH] = P[HTH] = P[HHT] = 0

P[TTH] = P[THT] = P[HTT] = P[HHH] = 1/4

My notation is that TTH means that coin A is tails, coin B tails, and coin C heads. Etc.
1. Are the individual coins fair (i.e., heads half the time)?
```
.











.
```
2. Are coins A and B independent?
```
.











.
```
3. Are all 3 coins independent?
```
.











.
```
This question is about a continuous probability distribution on 2 variables.

$$f_{XY}(x,y) = \begin{cases} c x y & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$

The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).

c is some constant, but I didn't tell you what it is.
1. What is c?
```
.











.
```
2. What is $F_{XY}(x,y)$?
```
.











.
```
3. What is $f_X(x)$?
```
.











.
```
4. Are X and Y independent?
```
.











.
```
5. What is $P[X\le Y]$ ?
```
.











.
```
6. Define a new random variable $Z=X+Y$. What is $F_Z(z)$?
```
.











.
```
7. What is $E[X]$?
```
.











.
```
8. What is $COV[X,Y]$?
```
.











.
```
9. What is $\rho_{X,Y}$?
```
.











.
```
10. What is $f_Y(y|x)$?
```
.











.
```
11. What is $E[Y|x]$?
```
.











.
```
Compute $$\int_0^\infty e^{-x^2} dx$$ .
```
.











.
```
What is the legal range for a correlation coefficient?
```
.











.
```
What is the variance of the sum of 100 independent variables, each of which is N(0,1)?
```
.











.
```
You have 10 independent random variables. Each is uniform on [0,1]. What is the expected value of the max?
```
.











.
```
You toss 10 independent fair coins, one after the other.
1. What is the expected total number of heads, given that the first 5 coins came up heads?
```
.











.
```
2. What is the probability of the total number of heads being 10, given that the first 5 coins came up heads?
```
.











.
```

End of exam 3, total 90 points (considering that 3 questions aren't graded).

Engineering Probability Exam 3 solution - Tues 2018-05-08

W Randolph Franklin (WRF), RPI — Mon, 07 May 2018 04:00:00 GMT

Name, RCSID: W. Randolph Franklin, frankwr

OK to give the formulas w/o working them out.

Rules:

You have 80 minutes.
You may bring three 2-sided 8.5"x11" papers with notes.
You may bring a calculator.
You may not share material with each other during the exam.
No collaboration or communication (except with the staff) is allowed.
Check that your copy of this test has all nine pages.
Each part of a question is worth 5 points.
You may cross out three question parts, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

You toss two coins. Each comes up heads half of the time. However, for some funny reason, they both come up heads together, or both come up tails together. Intuitively, they not independent. This question is to prove that from the definition of independence.

P[HT] = 0. However P[A=H] = P[B=T] = 1/2, so P[HT] != P[A=H]P[B=T]. That's the def of not independent.
This time, you toss three coins, A, B, and C. These are the probabilities:

P[TTT] = P[THH] = P[HTH] = P[HHT] = 0

P[TTH] = P[THT] = P[HTT] = P[HHH] = 1/4

My notation is that TTH means that coin A is tails, coin B tails, and coin C heads. Etc.
1. Are the individual coins fair (i.e., heads half the time)?
  
  P[A=H] = 0+0+1/4+1/4 = 1/2 so fair. Ditto B and C.
2. Are coins A and B independent?
  
  P[A=H,B=H] = 1/4, good. P[A=H,B=T]=1/4, good. P[TH] = 1/4, good. P[TT] = 1/4, good. Yes independent.
3. Are all 3 coins independent?
  
  P[HHH] = 1/4 != P[A=H]P[B=H]P[C=H] = 1/8. Not independent.
This question is about a continuous probability distribution on 2 variables.

$$f_{XY}(x,y) = \begin{cases} c x y & \text{ if } (0\le x) \ \& \ (0\le y)\ \& \ (0\le x+y \le 1) \\ 0 & \text{ otherwise}\end{cases}$$

The nonzero region is the triangle with vertices (0,0), (1,0) and (0,1).

c is some constant, but I didn't tell you what it is.
1. What is c?
  
  $\int_0^1\int_0^{1-x} xy\ dy\ dx = 1/24$ so $c=24$
2. What is $F_{XY}(x,y)$?
  
  $$F_{XY}(x,y)=\begin{cases} 0 & \text{ if } x\le0 \cup y\le0 \\ 1 & \text{ if } x\ge 1 \cap y\ge1 \\ 6x^2y^2 & \text{ if } 0\le x \cap 0\le y \cap x+y\le1 \\ (\int_0^x\int_0^{1-x} + \int_0^{1-y}\int_{1-x}^y + \int_{1-y}^x\int_{1-x}^{1-x_0}) (12x_0y_0 dy_0dx_0) & \text{ otherwise}\end{cases}$$
  
  The last case above splits the nonzero integration region into two rectangles and a triangle.
  
  It's also acceptable to draw a figure and say something intelligent w/o being explicit about all the details.
3. What is $f_X(x)$?
  
  $f_X(x)= \int_0^{1-x}f_{XY}(x,y) dy = 12x(1-x)^2$
  
  Note that $\int_0^1 f_X(x)=1$, which is correct.
4. Are X and Y independent?
  
  $f_X(x)=12x(1-x)^2,f_Y(y)=12y(1-y)^2,f_X(x)f_Y(y)\ne f_{XY}(x,y)$
  
  No.
5. What is $P[X\le Y]$ ?
  
  $\int_0^1\int_0^{\min(x,1-x)} f_{XY}(x,y) dy\ dx$. However, since $f_{XY}(x,y) = f_{XY}(y,x)$, $P[X\le Y]=1/2$
6. Define a new random variable $Z=X+Y$. What is $F_Z(z)$?
  
  $f_Z(z) = \int_0^z f_{XY}(x,z-x) dx = 24\int_0^z x(z-x)dx$ for $0\le z\le 1$
  
  $F_Z(z) = \int f_Z(z)dz$
7. What is $E[X]$?
  
  $\int_0^1 xf_X(x)dx = 2/5$
8. What is $COV[X,Y]$?
  
  E[XY] = $\int_0^1\int_0^{1-x}xyf_{XY}dx dy=8\int_0^1x^2(1-x)^4 dx, E[X]=E[Y]=2/5$, COV[X,Y]=E[XY]-E[X]E[Y]. You don't have to work through the math.
9. What is $\rho_{X,Y}$?
  
  $\sigma_X=\sigma_Y= E[X^2]-E[X]^2$
  
  $\rho_{X,Y}=COV[X,Y]/(\sigma_X\sigma_Y)$
10. What is $f_Y(y|x)$?
  
  $f_Y(y|x)=f(x,y)/f(x) = 12xy/(4x^3) = 2\frac{y}{x^2}$
11. What is $E[Y|x]$?
  
  Integrate the above over $y$ to get $x^{-2}$
Compute $$\int_0^\infty e^{-x^2} dx$$ .

Consider $\sigma=1/\sqrt{2}$. Working a little, this will give $\int_0^\infty e^{-x^2} dx=\sqrt{\pi}/2=0.886$. It was also ok just to use a calculator.
What is the legal range for a correlation coefficient?

-1 to 1
What is the variance of the sum of 100 independent variables, each of which is N(0,1)?

100.
You have 10 independent random variables. Each is uniform on [0,1]. What is the expected value of the max?

Let $W=\max(X_i). F_W(w)=w^{10}. f_W(w)=10w^9.E[W]=10/11.$
You toss 10 independent fair coins, one after the other.
1. What is the expected total number of heads, given that the first 5 coins came up heads?
  
  The last 5 coins do not depend on the first 5. So the expectation is 7.5.
2. What is the probability of the total number of heads being 10, given that the first 5 coins came up heads?
  
  1/32

End of exam 3, total 90 points (considering that 3 questions aren't graded).

Engineering Probability Class 19 and Exam 2 - Thu 2018-03-29

W Randolph Franklin (WRF), RPI — Thu, 29 Mar 2018 04:00:00 GMT

Name, RCSID:

.



.

Rules:

You have 80 minutes.
You may bring two 2-sided 8.5"x11" paper with notes.
You may bring a calculator.
You may not share material with each other during the exam.
No collaboration or communication (except with the staff) is allowed.
Check that your copy of this test has all eleven pages.
Each part of a question is worth 5 points.
You may cross out two questions, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

Consider this probability distribution:

$$f_X(x)= \begin{cases} a(2-x) & \text{if } 0\le x\le1\\ 0&\text{otherwise}\end{cases}$$
1. What is $a$?
```
.











.
```
2. What is $F_X(x)$?
```
.











.
```
3. What is E[X]?
```
.











.
```
4. What is the reliability, R[x]?
```
.











.
```
5. What is the MTTF?
```
.











.
```
6. What is the failure rate, r(x)?
```
.











.
```
7. What is $f_X(x|x>.5)$?
```
.











.
```
Define a new r.v. Y=2X, where X is the r.v. in the previous question.
1. What is $f_Y(y)$?
```
.











.
```
2. What is $F_Y(y)$?
```
.











.
```
3. What is E[Y]?
```
.











.
```
Your web server gets on the average 1 hit per second. The possible clients are independent of each other.
1. What is the name of appropriate distribution for the number of hits per second?
```
.











.
```
2. What is the probability that it gets exactly one hit in the next two seconds?
```
.











.
```
3. What is the name of appropriate probability distribution for the time between successive hits?
```
.











.
```
4. What is the probability that the time between two successive hits is less than two seconds?
```
.











.
```
Let X be an exponential random variable with mean 1.
1. Using the Markov inequality, what's P[X>3]?
```
.









.
```
2. Using the Chebyshev inequality, what's P[X>3]?
```
.








.
```
3. What's the exact P[X>3]?
```
.











.
```
Let X be a normal random variable with mean 100 and standard deviation 10. Give the following numbers, using the supplied table.
1. P[X>100].
```
.






.
```
2. P[80<X<100].
```
.






.
```
You're tossing 10000 fair coins. What's the probability of getting between 5000 and 5100 heads? Use the table.
```
.









.
```
Evaluate $$\int_0^\infty e^{-2 x^2} dx$$
```
.











.
```
Let $f_X(x) = 1$ and $f_Y(y)=2y$, both in the range $0\le x, y\le1$.

Let Z=max(X,Y).

What is E[Z]?
```
.














.
```

Normal distribution:

x          f(x)      F(x)      Q(x)
-3.0000    0.0044    0.0013    0.9987
-2.9000    0.0060    0.0019    0.9981
-2.8000    0.0079    0.0026    0.9974
-2.7000    0.0104    0.0035    0.9965
-2.6000    0.0136    0.0047    0.9953
-2.5000    0.0175    0.0062    0.9938
-2.4000    0.0224    0.0082    0.9918
-2.3000    0.0283    0.0107    0.9893
-2.2000    0.0355    0.0139    0.9861
-2.1000    0.0440    0.0179    0.9821
-2.0000    0.0540    0.0228    0.9772
-1.9000    0.0656    0.0287    0.9713
-1.8000    0.0790    0.0359    0.9641
-1.7000    0.0940    0.0446    0.9554
-1.6000    0.1109    0.0548    0.9452
-1.5000    0.1295    0.0668    0.9332
-1.4000    0.1497    0.0808    0.9192
-1.3000    0.1714    0.0968    0.9032
-1.2000    0.1942    0.1151    0.8849
-1.1000    0.2179    0.1357    0.8643
-1.0000    0.2420    0.1587    0.8413
-0.9000    0.2661    0.1841    0.8159
-0.8000    0.2897    0.2119    0.7881
-0.7000    0.3123    0.2420    0.7580
-0.6000    0.3332    0.2743    0.7257
-0.5000    0.3521    0.3085    0.6915
-0.4000    0.3683    0.3446    0.6554
-0.3000    0.3814    0.3821    0.6179
-0.2000    0.3910    0.4207    0.5793
-0.1000    0.3970    0.4602    0.5398

Normal distribution:

x          f(x)      F(x)      Q(x)
      0    0.3989    0.5000    0.5000
 0.1000    0.3970    0.5398    0.4602
 0.2000    0.3910    0.5793    0.4207
 0.3000    0.3814    0.6179    0.3821
 0.4000    0.3683    0.6554    0.3446
 0.5000    0.3521    0.6915    0.3085
 0.6000    0.3332    0.7257    0.2743
 0.7000    0.3123    0.7580    0.2420
 0.8000    0.2897    0.7881    0.2119
 0.9000    0.2661    0.8159    0.1841
 1.0000    0.2420    0.8413    0.1587
 1.1000    0.2179    0.8643    0.1357
 1.2000    0.1942    0.8849    0.1151
 1.3000    0.1714    0.9032    0.0968
 1.4000    0.1497    0.9192    0.0808
 1.5000    0.1295    0.9332    0.0668
 1.6000    0.1109    0.9452    0.0548
 1.7000    0.0940    0.9554    0.0446
 1.8000    0.0790    0.9641    0.0359
 1.9000    0.0656    0.9713    0.0287
 2.0000    0.0540    0.9772    0.0228
 2.1000    0.0440    0.9821    0.0179
 2.2000    0.0355    0.9861    0.0139
 2.3000    0.0283    0.9893    0.0107
 2.4000    0.0224    0.9918    0.0082
 2.5000    0.0175    0.9938    0.0062
 2.6000    0.0136    0.9953    0.0047
 2.7000    0.0104    0.9965    0.0035
 2.8000    0.0079    0.9974    0.0026
 2.9000    0.0060    0.9981    0.0019
 3.0000    0.0044    0.9987    0.0013

End of exam 1, total 100 points (considering that 2 questions aren't graded).

Engineering Probability Class 19 and Exam 2 Solution - Thu 2018-03-29

W Randolph Franklin (WRF), RPI — Thu, 29 Mar 2018 04:00:00 GMT

Name, RCSID:

WRF solution

Rules:

You have 80 minutes.
You may bring two 2-sided 8.5"x11" paper with notes.
You may bring a calculator.
You may not share material with each other during the exam.
No collaboration or communication (except with the staff) is allowed.
Check that your copy of this test has all eleven pages.
Each part of a question is worth 5 points.
You may cross out two questions, which will not be graded.
When answering a question, don't just state your answer, prove it.

Questions:

Consider this probability distribution:

$$f_X(x)= \begin{cases} a(2-x) & \text{if } 0\le x\le1\\ 0&\text{otherwise}\end{cases}$$
1. What is $a$?
  
  We require that $\int_0^1 f(x) = 1$.
  
  So, $a=2/3$. You don't have to write it, but this gives $f_X(x) = \frac{4}{3}-\frac{2}{3}x$ if $0<x<1$.
2. What is $F_X(x)$?
  
  $$F(x) = \int_0^x f(w) dw = \begin{cases} 0 & \text{if } x\le0 \\ \frac{4}{3}x - \frac{x^2}{3} & 0<x<1 \\ 1 & 1\le x \end{cases} $$
  
  You can use any placeholder variable (other than $x$) in place of $w$.
  
  It doesn't matter where you say $<$ versus $\le$.
3. What is E[X]?
  
  $$E[X] = \int_0^1 x f(x) dx = \int (4/3 x -2/3 x^2)dx = \left. \left(\frac{2 x^2}{3} - \frac{2 x^3}{9}\right)\right|_0^1 = \frac{4}{9}$$
4. What is the reliability, R[x]?
  
  $$R(x) = 1 - F(x) = \begin{cases} 1 - \frac{4}{3}x + \frac{x^2}{3} & 0<x<1 \\ 0 & 1\le x \end{cases} $$
  
  Lifetimes are nonnegative, so I deleted the case for $x<0$, but it doesn't matter.
5. What is the MTTF?
  
  MTTF = $$\int_0^1 R(x) dx = \int \left(1 - \frac{4}{3}x + \frac{x^2}{3}\right) dx = \frac{4}{9}$$
  
  MTTF=E[X]. The integral goes up to 1 because R(x) is 0 when x>1.
6. What is the failure rate, r(x)?
  
  For $0<x<1$,
  
  $$r(x) = \frac{-R'(x)}{R(x)} = \frac{\frac{4}{3}-\frac{2x}{3}}{1 - \frac{4}{3}x + \frac{x^2}{3}} = \frac{4-2x}{3 - 4 x + x^2}$$
  
  You don't need to simplify it.
  
  Note that $r(x)$ grows to infinity as $x$ approaches 1.
7. What is $f_X(x|x>.5)$?
  
  $P[x>.5] = 1-F(.5) = 5/12$.
  
  $$f_X(x|x>.5) = f(x)/P[x>.5] = \begin{cases} \frac{8}{5}(2-x) & \text{if } 0.5\le x\le1\\ 0&\text{otherwise}\end{cases}$$
  
  As a check, you can see that $\int f(x|x>.5) dx = 1$.
Define a new r.v. Y=2X, where X is the r.v. in the previous question.
1. What is $f_Y(y)$?
  
  The nonzero domain for $f_X(x)$ is $0<x<1$, and Y=2X.
  
  So the nonzero domain for $f_Y(y)$ will be $0<y<2$.
  
  $dy/dx = 2$, so
  
  $$f_Y(y) = f_X(y/2)/2 = \begin{cases} \frac{2}{3} - \frac{y}{6} & 0<y<2\\ 0 & \text{otherwise}\end{cases}$$
2. What is $F_Y(y)$?
  
  $$F_Y(y) = F_X(y/2)$$
  
  $$F_Y(y)= \begin{cases} \frac{2}{3} y - \frac{y^2}{12} & 0<y<2\\ 0 & y<0 \\ 1 & y>2 \end{cases}$$
3. What is E[Y]?
  
  $$ E[Y] = \int_0^2 y\left(\frac{2}{3} - \frac{y}{6}\right) dy = \left.\left(y^2/3-y^3/18\right)\right|_0^2 = \frac{8}{9}$$
  
  Alternatively, E[Y] = 2 E[X].
Your web server gets on the average 1 hit per second. The possible clients are independent of each other.
1. What is the name of appropriate distribution for the number of hits per second?
  
  Poisson
2. What is the probability that it gets exactly one hit in the next two seconds?
  
  That r.v. is Poisson with $\alpha=2$ so $$P[X=1] = \frac{2^1 e^{-2}}{1!} = 2 e^{-2} = .27$$
  
  full points for $2 e^{-2}$.
3. What is the name of appropriate probability distribution for the time between successive hits?
  
  Exponential
4. What is the probability that the time between two successive hits is less than two seconds?
  
  Mean: $1/\lambda= 1$. $F(x) = 1-e^{-x}$. $F(2) = 1-e^{-2}=.14=.86$.
  
  full points for $1-e^{-2}$.
Let X be an exponential random variable with mean 1.
1. Using the Markov inequality, what's P[X>3]?
  
  See page 181. $\mu=1$. $P\le 1/3$.
2. Using the Chebyshev inequality, what's P[X>3]?
  
  $\mu=\sigma=1$, P[X<0]=0, so $$P[X>3]= P[|X-1|>2] \le 1/4$$
3. What's the exact P[X>3]?
  
  $F(x) = 1-e^{-x}$ so $P[X>3] = 1-F[3] = e^{-3} = .05$
  
  full points for $e^{-3}$.
Let X be a normal random variable with mean 100 and standard deviation 10. Give the following numbers, using the supplied table.
1. P[X>100].
  
  0.5
2. P[80<X<100].
  
  Converted to $\mu=0,\ \sigma=1$, this is P[-2<Y<0] = F(0)-F(-2) = .5 - .02 = .48.
You're tossing 10000 fair coins. What's the probability of getting between 5000 and 5100 heads? Use the table.

This is a Bernoulli r.v. with $\mu=5000,\ \sigma=\sqrt{npq}=50$.

Use a normal approximation; the answer is F(2)-F(0) = .98-.5 = .48.
Evaluate $$\int_0^\infty e^{-2 x^2} dx$$

For $\mu=0$, what value of $\sigma$ would make $f(x) = c e^{-2 x^2}$?

$$f(x) = \frac{1}{\sqrt{2\pi} \cdot \sigma} \exp\left(\frac{-x^2}{2\sigma^2}\right)$$

so let $\sigma=1/2$ and

$$f(x) = \sqrt{\frac{2}{\pi} } e^{\left(-2 x^2\right)}$$

So $$\int_\infty^\infty e^{-2 x^2} = \sqrt{\frac{\pi}{2} }$$

and $$\int_0^\infty e^{-2 x^2} dx$$ = $$\sqrt{\frac{\pi}{8} }$$

which could be written various ways.
Let $f_X(x) = 1$ and $f_Y(y)=2y$, both in the range $0\le x, y\le1$.

Let Z=max(X,Y).

What is E[Z]?

$F_X(x) = \int f_X(x) dx = x$,

$F_Y(y) = y^2$.

$F_Z(z) = F_X(z) F_Y(z) = z^3$,

$f_Z(z) = 3 z^2$,

$$E[Z] = \int_0^1 3 z^3 dz = 3/4 z^4|_0^1 = \frac{3}{4}$$.

Normal distribution:

x          f(x)      F(x)      Q(x)
-3.0000    0.0044    0.0013    0.9987
-2.9000    0.0060    0.0019    0.9981
-2.8000    0.0079    0.0026    0.9974
-2.7000    0.0104    0.0035    0.9965
-2.6000    0.0136    0.0047    0.9953
-2.5000    0.0175    0.0062    0.9938
-2.4000    0.0224    0.0082    0.9918
-2.3000    0.0283    0.0107    0.9893
-2.2000    0.0355    0.0139    0.9861
-2.1000    0.0440    0.0179    0.9821
-2.0000    0.0540    0.0228    0.9772
-1.9000    0.0656    0.0287    0.9713
-1.8000    0.0790    0.0359    0.9641
-1.7000    0.0940    0.0446    0.9554
-1.6000    0.1109    0.0548    0.9452
-1.5000    0.1295    0.0668    0.9332
-1.4000    0.1497    0.0808    0.9192
-1.3000    0.1714    0.0968    0.9032
-1.2000    0.1942    0.1151    0.8849
-1.1000    0.2179    0.1357    0.8643
-1.0000    0.2420    0.1587    0.8413
-0.9000    0.2661    0.1841    0.8159
-0.8000    0.2897    0.2119    0.7881
-0.7000    0.3123    0.2420    0.7580
-0.6000    0.3332    0.2743    0.7257
-0.5000    0.3521    0.3085    0.6915
-0.4000    0.3683    0.3446    0.6554
-0.3000    0.3814    0.3821    0.6179
-0.2000    0.3910    0.4207    0.5793
-0.1000    0.3970    0.4602    0.5398

Normal distribution:

x          f(x)      F(x)      Q(x)
      0    0.3989    0.5000    0.5000
 0.1000    0.3970    0.5398    0.4602
 0.2000    0.3910    0.5793    0.4207
 0.3000    0.3814    0.6179    0.3821
 0.4000    0.3683    0.6554    0.3446
 0.5000    0.3521    0.6915    0.3085
 0.6000    0.3332    0.7257    0.2743
 0.7000    0.3123    0.7580    0.2420
 0.8000    0.2897    0.7881    0.2119
 0.9000    0.2661    0.8159    0.1841
 1.0000    0.2420    0.8413    0.1587
 1.1000    0.2179    0.8643    0.1357
 1.2000    0.1942    0.8849    0.1151
 1.3000    0.1714    0.9032    0.0968
 1.4000    0.1497    0.9192    0.0808
 1.5000    0.1295    0.9332    0.0668
 1.6000    0.1109    0.9452    0.0548
 1.7000    0.0940    0.9554    0.0446
 1.8000    0.0790    0.9641    0.0359
 1.9000    0.0656    0.9713    0.0287
 2.0000    0.0540    0.9772    0.0228
 2.1000    0.0440    0.9821    0.0179
 2.2000    0.0355    0.9861    0.0139
 2.3000    0.0283    0.9893    0.0107
 2.4000    0.0224    0.9918    0.0082
 2.5000    0.0175    0.9938    0.0062
 2.6000    0.0136    0.9953    0.0047
 2.7000    0.0104    0.9965    0.0035
 2.8000    0.0079    0.9974    0.0026
 2.9000    0.0060    0.9981    0.0019
 3.0000    0.0044    0.9987    0.0013

End of exam 1, total 100 points (considering that 2 questions aren't graded).

Engineering Probability Class 12 and Exam 1 Solution - Mon 2018-02-26

W Randolph Franklin (WRF), RPI — Sun, 25 Feb 2018 05:00:00 GMT

Name, RCSID: WRF solutions

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

Rules:

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

Questions:

(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.
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 .
(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.
(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.
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.
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.
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 = $\sqrt{5}$
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.

Engineering Probability Class 12 and Exam 1 - Mon 2018-02-26

W Randolph Franklin (WRF), RPI — Sat, 24 Feb 2018 05:00:00 GMT

Name, RCSID:

.




.

Rules:

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

Questions:

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






.
```
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?
```
.










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














.
```
(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?
```
.















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












.
```
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?
```
.





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







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







.
```
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?
```
.











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











.
```
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?
```
.











.
```
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.
```
.











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











.
```
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.
```
.











.
```
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.
```
.











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











.
```
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?
```
.











.
```

End of exam 1, total 70 points.