University Solutions to Final Exam ECE 313
of Illinois Page 1 of 3 Spring 2003
1.(a) There are ( )
6
3 = 6×5×4
1×2×3 = 20 short lists of which ( )
5
2 = 5×4
1×2 = 10 include Chuck, and thus Chuck has
probability 10/20 = 1/2 of being on the short list. By symmetry, everyone has probability 1/2 of being on
the short list.
(b) There is only one short list {Andy, Chuck, Eddie} that has no women on it. Hence, the probability is
19/20 that there is at least one woman on the short list.
(c) Chuck is selected as monarch with conditional probability 1/3 if he is on one of the three short lists that
include him and two of Andy, Eddie and Fergie, and he is selected as monarch with probability 1/2 if he is
on of the three short lists that include him and Di and one of Andy, Eddie and Fergie. Hence, from the
theorem of total probability we get that P{Chuck selected as monarch} = 1
3×3
20 + 1
2×3
20 = 1
8. More directly,
since Beth is on the short list with probability 1/2, she is selected as monarch with probability 1/2 also.
By symmetry, the other 4 nonDians are equally likely to be selected if Beth is not, and thus are selected
with probability 1/8 each.
(d) P{Di on list | Chuck selected} = P{Chuck selected and Di on list}
P{Chuck selected}
= P{Chuck selected | C, D and one of A, E, F on list}P{C, D and one of A, E, F on list}
P{Chuck selected}
= (1/2)(3/20)
(1/8) = 3
5.
(e) The other two women must be Di and Fergie (not Beth, since Chuck was selected), and hence from part (d)
we get that P{two women on list | Chuck selected} = 1/5.
2. P{HHH} = P{TTT} = 1/8. Thus, a decision is made with probability 3/4 and re-tossing is required with
probability 1/4. The number of rounds X is thus a geometric random variable with parameter p = 3/4.
Note (from symmetry) that Groucho, Chico and Harpo have equal probability 1/3 of winning the game.
(a) P{at least 3 rounds required} = P{X > 2} = q2 = (1/4)2 = 1/16.
(b) P{X is even} = pq + pq3 + pq5 + … = pq
1–q2 = q
1+q = 1/4
1+1/4 = 1
5.
(c) P{X > 2 | X is even} = P{X > 2 and X is even}
P{X is even} = pq3 + pq5 + …
1/5 = 1/5 – 3/16
1/5 = 1 – 15
16 = 1
16.
(d) Now the outcomes are {HHH, THH, HHT, THT} so that a decision is still made with probability 3/4 and
re-tossing is required with probability 1/4. Each player still has the same probability 1/3 of winning.
(e) Now the outcomes are {HHT, THT} so that either Chico or Harpo wins (with probability 1/2) on the very
first round. So, Harpo actually helps Chico by his strategy. Note that if Harpo had also substituted a two-
headed coin, then Groucho wins with probability 1, though not necessarily on the first round!
3. Since the area under the pdf must be 1, we readily get that a + b = 2.
(a) The variance is a measure of the spread of the mass, and therefore is smallest when either a = 0 and b = 2 or
when a = 2 and b = 0. In either case, the pdf is a uniform density on an interval of length 1/2, and hence
the minimum variance is (1/2)2/12 = 1/48.
More analytically, ∫
0
1/2
u du = 1
8 and ∫
1/2
1
u du = 3
8 from which we get that E[X] = a + 3b
8 = 3 – a
4.
Similarly, ∫
0
1/2
u2 du = 1
24 and ∫
1/2
1
u2 du = 7
24 from which we get that E[ X2] = a + 7b
24 = 7 – 3a
12 .
Hence, var(X) = E[X2] – (E[X])2 = 7 – 3a
12 – 9 – 6a + a2
16 = 1 + 6a –3a2
48 . As a function of a, this
quadratic has minimum value 1/48 on the interval [0, 2] , and the minimum value occurs at a = 0 and a = 2.
Those who set the derivative to 0 and got the extremum at a = 1 actually found the maximum variance
1/12, not the minimum variance! The second derivative test, if tried, would have revealed the error.
(b) FX(3/4) = a
2 + b
4 = a + 2
4.
(c) If E[X] = 3 – a
4 = 5
8, we readily get a = 1
2 and hence FX(3/4) = a + 2
4 = 5
8.
