University of Illinois Spring 2009
ECE 313: Hour Exam I
Monday March 2, 2009
7:00 p.m. — 8:00 p.m.
100 Noyes Laboratory
1. (a) The hat contains balls {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}with equal probability
1
6. Hence, P(3 ball is in the hat) = 1
2.
(b) From the answer to part (a), we can read off the values of the pmf:
pX(3) = 1
6,pX(4) = 1
6,pX(5) = 2
6=1
3,pX(6) = 1
6,pX(3) = 1
6, and pX(u) = 0 for all other
values of u. Sanity check: pX(3) + pX(4) + pX(5) + pX(6) + pX(7) = 1 as it should be.
(c) Following the theft of the 3 ball, the hat contains {1,2},{1},{1,4},{2},{2,4},{4}
with equal probability 1
6. Hence, by the law of total probability,
P(ball drawn has an odd number) =1
2+1+1
2+0+0+0×1
6=1
3.
2. (a) P(Fred wins game) = P{H, T T H, T T T T H, . . .}=P{H}+P{TTH}+P{TTTTH}+
· · · =1
2+1
221
2+1
241
2+· · · =1
21 + 1
4+1
42
+· · · =1
2×1
1−1/4=4
2×3=2
3.
Similarly, P(Wilma wins first game) = 1
3.
Alternatively, with Fand Wdenoting the events that Fred and Wilma respectively win
the first game, and Hthe event that the first toss (by Fred) results in a Head, we have
P(F|H) = 1 and P(W|H) = 0). Also, if T=Hcoccurs on the first toss, then Wilma
gets to toss in a brand new game and thus P(F|T) = P(W), P(W|T) = P(F). By
the law of total probability, we get
P(F) = P(F|H)P(H) + P(F|T)P(T) = 1 ×1
2+P(W)×1
2=1
2+P(W)
2,
P(W) = P(W|H)P(H) + P(W|T)P(T)=0×1
2+P(F)×1
2=P(F)
2.
It follows that P(F) = 1
2+1
4×P(F) and hence P(F) = 2
3and P(W) = 1
3.
(b) Let Fand Wdenote the events that Fred and Wilma respectively win a game, and let
FFW denote the event that Fred wins the first two games and Wilma the third, etc.
Remembering that the winner of a game has only a 1
3chance of winning the next game,
we proceed to calculate systematically as follows:
Result Value of XProbability Result Value of XProbability
FFF −32
3×1
3×1
3=2
27 WWF +1 1
3×1
3×2
3=2
27
FFW −12
3×1
3×2
3=4
27 WFW +1 1
3×2
3×2
3=4
27
FWF −12
3×2
3×2
3=8
27 FWW +1 2
3×2
3×1
3=4
27
WFF −11
3×2
3×1
3=2
27 WWW +3 1
3×1
3×1
3=1
27
