University Problem Set #4: Solutions ECE 313
of Illinois Page 1 of 3 Summer 2003
5. Let A2, A3, . . . , A12 denote the events that 2, 3, 4, . . . , 12 is rolled with two fair dice.
(a) P(winning on first roll) = P(A7∪A11) = P(A7) + P(A11) = 6/36 + 2/36 = 2/9.
P(losing on the first roll) = P(A2∪A3∪A12) = 1/36 + 2/36 + 1/36 = 1/9.
(b) P(shooter’s point is 4) = P(A4) = 3/36 = 1/12 = P(A10) = P(shooter’s point is 10).
P(shooter’s point is 5) = P(A5) = 4/36 = 1/9 = P(A9) = P(shooter’s point is 9).
P(shooter’s point is 6) = P(A6) = 5/36 = P(A8) = P(shooter’s point is 8).
(c) For disjoint events A and B, P(event A occurs before B) = P(A)/(P(A) + P(B)). Hence,
P(shooter makes point|shooter’s point is 4) = 3/(3+6) = 1/3 = P(shooter makes point|shooter’s point is 10)
P(shooter makes point|shooter’s point is 5) = 4/(4+6) = 2/5 = P(shooter makes point|shooter’s point is 9)
P(shooter makes point|shooter’s point is 6) = 5/(5+6) = 5/11 = P(shooter makes point|shooter’s point is 8)
(d) P(winning at craps) = P(winning on first roll)+ ∑
i
P(making point|point is i)P(shooter’s point is i)
= (8/36) + 2×[(3/36)×(1/3) + (4/36)×(2/5) + (5/36)×(5/11)] = (8/36) + (2/36)×[1 + 8/5 + 25/11]
= (8/36) + (2/36)×[55 + 88 + 125]/55 = (440 + 536)/(36×55) = 244/495 = 0.49292929…
(e) P(making 8 the hard way) = P(rolling 4-4 before rolling 7 or rolling 8 the easy way)
= (1/36)/(1/36 + (6/36 + 4/36)) = 1/11. Thus, in a long sequence of trials when the shooter’s point is 8,
one-eleventh of the time you win $10 while ten-eleventh of the time you lose your $1. Thus, in the long
run, you neither make nor lose money on the average, and the bet is perfectly fair.
6.(a) The game winner wins the last point in the game, i.e., a score of 4-x or x-4 (with 0 ≤ x < 3) means that the
x points won by the loser must have been among the first 3+x points. Thus,
P(4-x) = ()
3+x
x p3qxp, P(x-4) = ()
3+x
x q3pxq, giving P(4-0) = p4; P(4-1) = 4p4q; P(4-2) = 10p4q2;
P(0-4) = q
4; P(1-4) = 4pq4; P(2-4) = 10p2q4.
(b) Deuce is reached with probability ()
6
3 p3q3 = 20p3q3. To show that the sum is 1, just multiply it all out
and hope you don't make an error (or use Mathematica).
(c) P(A wins next two points|deuce) = p2. P(B wins next two points|deuce) = q2.
P(each wins one point|deuce) = P(deuce again) = 2pq (why 2 ?)
(d) P(A wins the game|deuce) = p2 + (2pq)p2 + (2pq)2p2 + … = p2/(1 – 2pq). Similarly,
P(B wins the game|deuce) = q2 + (2pq)q2 + (2pq)2q2 + … = q2/(1-2pq). As a sanity check, we note that
1 = (p+q)2 = p2 + q2 + 2pq so that 1 – 2pq = p2+q2 implying that p2/(1-2pq) + q2/(1-2pq) = 1, i.e. one of A
and B does ultimately win the game.
7.(a) The bill passes if the Conservative Republicans and at least one other group votes for it, i.e. the event
A∩(B∪C∪D) occurs, or if all groups except the Conservative Republicans vote for it, i.e. the event
Ac∩B∩C∩D occurs. Note that these are disjoint events (Why?). Now, (B∪C∪D)c = Bc∩Cc∩Dc by
DeMorgan's theorem, and thus we can partition A into two disjoint events as follows:
A = A∩(B∪C∪D) ∪ A∩Bc∩Cc∩Dc. Hence, we see that
P(A∩(B∪C∪D)) = P(A) – P(A∩Bc∩Cc∩Dc) = P(A) – P(A)P(Bc)P(Cc)(Dc) by independence. Thus,
P(bill passes) = P(A∩(B∪C∪D)) + P(Ac∩B∩C∩D) = P(A) – P(A)P(Bc)P(Cc)P(Dc) +
P(Ac)P(B)P(C)P(D) (independence again!) = 0.9×[1 – 0.4×0.5×0.8] + 0.1×0.6×0.5×0.2 = 0.756 + 0.006 =
0.762.
(b) The Conservative Republicans must be in favor for the motion to override to pass. Furthermore, since
politics makes strange bedfellows, either the Liberal Democrats must also be in favor or both the other
groups must also be in favor. Hence,
P(motion to override passes) = P[E(H ∪ FG)] = P(E)×P(H ∪ FG) by independence of E and H ∪ FG)
= P(E)×[P(H)+P(FG) – P(FGH)] = P(E)×[P(H) + P(F)P(G) – P(F)P(G)P(H)] (by independence again!)
= 0.99×[0.1 + 0.4×0.6 – 0.6×0.4×0.1] = 0.31284.
6.(a) With N-ary replication of each component, each replicated component has probability of failure pN since
all the N copies of the component must fail for the replicated component to fail. The probability that all M
replicated components are in working order is thus (1 – pN)M, and the probability of failure for the system
