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University of Illinois Spring 1998
ECE 313: Solutions to Problem Set
1. Let us first consider a more general situation. Suppose that event A occurs with probability p = P (A) on a single trial. If A occurs, we will call the trial a success, otherwise we call it a failure. Suppose that the
independent trials are repeated indefinitely, and let An;m denote the event that n successes occurbefore the
occurrence of m failures. Then An;m occurs if and only if there are at least n successes among the first
(n + m 1) trials. Indeed, if there are at least n successes among the first (n + m 1) trials, then there are at
most m 1 failures; thus n successes would occur before m failures. On the other hand, if there are at most
n 1 successes among the first (n + m 1) trials, then there are at least m failures; thus n successes would
not occur before m^ failures. The probability of at least^ n^ successes in^ (n^ +^ m^ ^ 1)^ trials is given by:
P (An;m ) =
n+Xm 1
i=n
n + m 1
i
pi^ (1 p)n+m ^1 i^ (1)
as shown in class. We will now apply this result to our problem. In our case, the trials consist of rolling a pair of fair dice. Notice, however, that wedo not care about those trials where neither 7 nor an even number is
rolled. Thus let us call a trialrelevant if either 7 or an even number result as the outcome. If S is an event that
a seven is rolled on a single trial and E is an event that an even number is rolled on a single trial, then
P (S ) =
and P (E ) =
Since these events are disjoint, the probability of the event R that a given trial is relevant is P (R) = P (E [
S ) = P (E ) + P (S ) = 2 = 3. We will call a relevant trial a success if 7 is rolled, otherwise we will call it a
failure. The probability of success is then given by
p = P (S jR) =
P (RjS )P (S )
P (R)
where we have used Bayes inversion formula to find P (S jR). The answer to the question in our problem can
be now obtained from (1) by substituting p = 0 : 25 , n = 2 , and m = 6. Thus
P (A 2 ; 6 ) =
X^7
i=
i
(0:25)i(0:75)^7 i^ = 1
X^1
i=
i
(0:25)i^ (0:75)^7 i
is desired the probability of rolling 7 twice before rolling an even number six times. (Exercise: where does
the second equality above come from?) Evaluating this expression we find that P (A 2 ; 6 ) = 1 36 10 = 47 =
2. The following table is quite helpful in solving this problem:
! in : : : AB C AB C c^ AB c^ C AB c^ C c^ Ac^ B C Ac^ B C c^ Ac^ B c^ C Ac^ B c^ C c
value of X (! ) 2 3 0 1 1 2 1 0
probability 0 : 125 0 : 125 0 : 125 0 : 125 0 : 125 0 : 125 0 : 125 0 : 125
( a) Referring to the table above, the random variable X takes on the values 1 ; 0 ; 1 ; 2 ; 3.
( b) Using the table again, we see that the probability mass function of X is given by pX ( 1) = pX (3) =
0 : 125 , pX (0) = pX (1) = pX (2) = 0 : 25 , and pX (u) = 0 for all u not in the set f 1 ; 0 ; 1 ; 2 ; 3 g. The
CDF of X can be easily found from its pmf, as follows
FX (u) =
0 u < 1
0 : 125 1 u < 0
0 : 375 0 u < 1
0 : 625 1 u < 2
0 : 875 2 u < 3
1 u 3
3. One possible probability measure on = f 1 ; 2 ; 3 ; 4 ; 5 g that would satisfy the requirements of parts ( a ) and ( b ) is defined by
P (f 1 g) = 0 : 1 ; P (f 2 g) = 0 : 1 ; P (f 3 g) = 0 : 2 ; P (f 4 g) = 0 : 2 ; P (f 5 g) = 0 : 4 ;
The solution in what follows is with respect to this probability measure.
( a) We can now define the random variable X as follows
X (i) = i for i = 1 ; 2 ; 3 ; 4 ; 5 (2)
It is immediate from the definition that X takes the values 1 ; 2 ; 3 ; 4 ; 5 with probabilities 0 : 1 ; 0 : 1 ; 0 : 2 ; 0 : 2 , 0 : 4 , as desired. Notice that there are other ways to define the random variable X to satisfy this property. For instance, we could define
X (1) = 2 ; X (2) = 1 ; X (i) = i for i = 3 ; 4 ; 5 (3)
( b) The random variable Y could be defined as follows
Y (3) =
p 2 ; Y (1) = Y (4) =
p 3 ; Y (2) = Y (5) = (4)
Again, this definition of Y satisfies the required property, but it is not unique. Why? ( c) It is easy to see that the random variable Z = X Y takes 5 different values, depending on the five possible outcomes in. For X and Y defined in (2) and (4), the possible values of Z are p 3 ; 2 ; 3
p 2 ; 4
p 3 ; 5
taken with probabilities 0 : 1 ; 0 : 1 ; 0 : 2 ; 0 : 2 ; 0 : 4 respectively. If the random variables X ; Y were defined differently, then the answer could be different. For instance, if X is defined as in (3) then Z takes the values 2
p 3 ; ; 3
p 2 ; 4
p 3 ; 5 .
4. ( a) Here FX (u) is not nondecreasing (FX (1) = 1 ; FX (2) = 0 ), and hence not a valid CDF.
( b) Here FX (u) is a valid CDF. It is continuous everywhere except at u = 0 , where it is right-continuous. We have P fjX j > 0 : 5 g = P fX < 0 : 5 g + P fX > 0 : 5 g and
P fX < 0 : 5 g + P fX > 0 : 5 g = FX ( 0 : 5 ) + [1 FX (0:5)] = 0 : 5 e ^1 + 0 : 25 e ^1 :^5
( c) Here FX (u) is not right-continuous at u = 0 , and therefore not a valid CDF.
−2^0 −1 0 1 2 3 4 5
1
u −−>
CDF of Random Variable X
Figure 1: Plot of CDF of the random variable X. Notice how all the properties of a valid CDF are satisfied.
5. ( a) The probability mass at any point is the height of the discontinuity or step (if one exists) at that point. From Figure 1 we see that
P fX = 1 g =
; P fX = 2 g =
; P fX = 3 g =
:
The total probability of the desired event, and hence P fY = ig is
P fY = ig =
2 i 2
2 2(i 1)
22 2 i
12 2 i
2 i 2
22 2 i
; i = 1 ; 2 ; : : : ; 6
0 ; otherwise
Those that have the enthusiasm can check and see that the above formula can be obtained by using expressions
for P fX = 2 i 1 g and P fX = 2 ig from Eq.(5) and plugging them in Eq.(6). The pmfs are tabulated below
(the pmf of X is zero after i = 11 , and the pmf of Y is zero after i = 6 ):
i 1 2 3 4 5 6
P fX = ig 0 0.0526 0.1053 0.1486 0.1734 0.
P fY = ig 0.0526 0.2539 0.3468 0.2553 0.0859 0.
i 7 8 9 10 11 12 ; : : :
P fX = ig 0.1486 0.1067 0.061 0.0249 0.0055 0
7. ( a) If the largest number selected out of n balls chosen is k , then P fY = k g is simply the probability that
n 1 balls were selected from numbers 1 ; 2 ; : : : ; (k 1) and one ball is number k. Clearly k n. So
P fY = k g =
k 1
n 1
N n
; k = n; : : : ; N
0 ; otherwise
( b) When we have selection with replacement we use a slightly different strategy to calculate P fY = k g.
Note that P fY = k g = P fY k g P fY k 1 g. The probability P fY k g is simply the
probability that the maximum number picked is less than or equal to k , which is the probability that the
first m balls picked were from numbers 1 ; 2 ; : : : ; k. Therefore P fY k g = k m^ =nm^. This gives
P fY = k g = P fY k g P fY k 1 g =
k m^ (k 1)m
nm^
; k = 1 ; : : : ; n
8. [Extra Credit] This is not a straightforward problem. And there are probably many ways to solve it. Here is one way: Let us denote P (A) and P (B ) by a and b respectively. Since A and B are independent, P (AB ) =
ab; P (Ac^ B c^ ) = (1 a)(1 b) and P (A B ) = a(1 b) + b(1 a). Let us further denote ab and a + b by
P and S respectively. We can rewrite P (AB ); P (Ac^ B c^ ) and P (A B ) in terms of S and P as:
P (AB ) = P ; P (Ac^ B c^ ) = 1 S + P ; P (A B ) = S 2 P :
Now we have to show that at least one of P , 1 S + P , and S 2 P is 4 = 9. Let us assume the contrary
and try to find a contradiction. So assume that
P < 4 = 9
1 S + P < 4 = 9
S 2 P < 4 = 9
P < 4 = 9
S P > 5 = 9
S 2 P < 4 = 9
The above three inequalities together describe a region in the S P plane bounded by the 3 straight lines:
P = 49 ; S P = 59 ; S 2 P = 49. This region is a triangle and is shown in Figure 2; it’s three vertices can
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 −
−0.
0
1
(1,4/9)
(4/3,4/9)
(2/3,1/9)
P = 4/
S−2P = 4/
S−P = 5/
S = 4P
2
....................................................^ ................. ........................................................ .....................................
........................ .....
Figure 2: Regions of interest in Problem 8.
easily be found to be
;
;
and
;
. Thus, any solution to the three inequalities above has to
be in the interior of this triangle.
However, this is not the end of the story. There is another inequality that S and P have to satisfy and that
is S 2 4 P. This is because the arithmetic mean of two numbers is always greater than or equal to their
geometric mean. Applying this to a and b we get a+ 2 b
p
ab ) (a + b)^2 4 ab, or S 2 4 P. The curve
S 2 = 4 P is also plotted in Figure 2, and the region S 2 4 P is the region below the curve. This curve passes
through two vertices of the triangle of interest: (4= 3 ; 4 =9) and (2= 3 ; 1 =9). However, no point in the triangle is
in the region described by S 2 4 P. Actually you could have found this out without having to plot the curves
by using some simple plane geometry.
Therefore there is no solution to the inequalities described by Equation (7) and S 2 4 P. This is the contra-
diction we were looking for, and hence at least one of P (AB ); P (Ac^ B c^ ); P (A B ) is greater than or equal
to 4 = 9.
