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University of Illinois Spring 1998
ECE 313: Solutions to Problem Set
1. ( a) Let Ai , respectively Bi , denote the event that the person who ranked i-th on the exam is a girl, respectively a boy. Since all the 10! rankings are equally likely, we obtain:
pX (1) = P (X =1) = P (A 1 ) =
pX (2) = P (X =2) = P (B 1 A 2 ) = P (A 2 jB 1 )P (B 1 ) =
pX (3) = P (X =3) = P (B 1 B 2 A 3 ) = P (A 3 jB 2 B 1 )P (B 2 jB 1 )P (B 1 ) =
where we have used the product rule for conditional probabilities. Continuing in this man- ner, we can further compute:
pX (4) = P (X = 4) = P (B 1 B 2 B 3 A 4 ) =
pX (5) = P (X = 5) = P (B 1 B 2 B 3 B 4 A 5 ) =
pX (6) = P (X = 6) = P (B 1 B 2 B 3 B 4 B 5 ) =
It is obvious from the statement of the problem that these are all the possible values of X. Indeed, we can also verify that pX (1) + pX (2) + + pX (6) = 1. ( b) The expected value of X can be easily computed directly by the definition, as the sum of six terms. We have:
E [X ] =
X^6
i=
i pX (i) = 1
To find the variance of X , we will also need to compute E [X 2 ]. This can be computed in a similar fashion, as follows:
E [X 2 ] =
X^6
i=
i^2 pX (i) = 1
Thus the variance of X is given by
Var[X ] = E [X 2 ] (E [X ])^2 =
2. Since the balls are placed back in the box after each round, the game consists of independent trials of the same experiment. The probability of success on a single trial, namely the probability of drawing exactly two white balls and two black balls, is given by
p =
2
2
4
( a) It is important to realize that the number of rounds X is a geometric random variable, with parameter p computed above. The expected value and variance of a geometric random variable were computed in class and are given by Ross, on page 183. Thus
E [X ] =
p
and Var[X ] =
p^2
p
( b) The key observation here is that Y = e X^. Thus the easiest way to find the expected value of the reward Y is by using LOTUS as follows:
E [Y ] = E [e X^ ] =
X^1
n=
e n^ pX (n) =
X^1
n=
e n^ (1 p)n ^1 p
where we have used the fact that pX (n) = (1 p)n ^1 p for all n 1. Indeed, this is the probability mass function of a geometric random variable, as discussed in class. The sum above can be easily computed as follows:
E [Y ] =
X^1
n=
e n^ (1 p)n ^1 p = pe ^1
X^1
n=
(1 p)e ^1
n 1
pe ^1 1 (1 p)e ^1
where we used the well-known expression for the sum of a geometric series. Evaluating the above expression for p = 18 = 35 , we obtain E [Y ] ' 0 : 230356.
3. ( a) The value c = 3 can be easily computed using the fact that
R 1
1^ fX^ (u)^ du^ =^1 for any probability density function. Thus
Z 1
fX (u) du = c
Z 1
0
(1 u)^2 du = c
Z 1
0
x^2 dx =
c 3
( b) We can find both E [X ] and E [X 2 ] by evaluating simple integrals, as follows:
E [X ] =
Z 1
u fX (u) du = 3
Z 1
0
u (1 u)^2 du = 0 : 25
E [X 2 ] =
Z 1
u^2 fX (u) du = 3
Z 1
0
u^2 (1 u)^2 du = 0 : 1
We can now compute the variance of X follows:
Var[X ] = E [X 2 ] (E [X ])^2 = 0 : 1 (0:25)^2 = 0 : 0375 =
( c) The idea in both cases is to express the probability that X satisfies the desired inequality as the probability that X belongs to an interval, or a union of intervals, of the real line. The latter kind of probability can then be computed by integration of the probability density function. Thus
P (6X 2 > 5 X + 1) = P ((6X + 1)(X 1) > 0) = P (fX < ^1 = 6 g [ fX > 1 g) where we had to solve a quadratic equation to find the factorization of 6 X 2 5 X 1. Now it is easy to see that the latter probability is zero, since fX (u) = 0 for u 62 (0; 1). Similarly, we solve a quadratic equation to find that
P (6X 2 > 7 X 2) = P ((3X 2)(2X 1) > 0) = P (fX < 1 = 2 g [ fX > 2 = 3 g)
( b) Using the result of part ( a ), the maximum allowable value of that would achieve the CEO’s goal is the largest that satisfies the inequality:
2
= 2[1 (0: 005 = )] 0 : 01
Referring to the tables of the standard Gaussian CDF, we find that the above inequality is satisfied for all 0 : 0019. ( c) This part is asking for P (fX < 0 : 896 g [ fX > 0 : 908 g). This probability can be computed as follows: 1 [FX (0:908) FX (0:896)] ' 1 [(2:67) ( 1 :33)] = 2 (2:67) (1:33) Referring to the tables of the standard Gaussian CDF, we find that the above is approxi- mately equal to 0 : 0956. Thus about 9 :56% of the screws will be defective.
6. ( a) Using the general properties of mean and variance, we find that:
E [2X +5] = 2 E [X ] + 5 = 3 Var[2X +5] = 22 Var[X ] = 16 Notice that there is no need to use the fact that X is Gaussian in this part.
The key idea in the rest of this problem is to define the random variable Y = X^ = X^2 +1 , and notice that Y is a standard Gaussian random variable. Thus
P (X < 0) = P
X +
2 <^
1 2
P ( 10 < X < 5) = P
2 <^
X + 2 <^
5+ 2
P (jX j > 5) = P (X > 5) + P (X < 5) = P
X +
2 >^
5+ 2
+ P
X +
2 <^
1 5 2
P (X 2 3 X + 2 < 0) = P ((X 1)(X 2) < 0) = P
2 <^
X + 2 <^
2+ 2
(Why is the last equality above true?) Now, we can express the probabilities at hand in terms of the random variable Y , as follows:
P (X < 0) = P (Y < 0 :5) P ( 10 < X < 5) = P ( 4 : 5 < Y < 3) P (jX j > 5) = P (Y > 3) + P (Y < 2) P (X 2 3 X + 2 < 0) = P (1 < Y < 1 :5) The identities ( x) = 1 (x) = Q(x) and Q( x) = 1 Q(x) = (x) are now used to manipulate answers into the required form, appropriate for Calculators A and B. P (Y < 0 :5) = (0:5) = 1 Q(0:5) P ( 4 : 5 < Y < 3) = (3) + (4:5) 1 = 1 Q(3) Q(4:5) P (Y > 3) + P (Y < 2) = 1 (3) + 1 (2) = Q(3) + Q(2) P (1 < Y < 1 :5) = (1:5) (1) = Q(1) Q(1:5) These are the required answers, appropriate for calculation using the two Calculators.
