University Problem Set #7: Solutions ECE 313
of Illinois Page 1 of 3 Spring 2003
1. There are ( )
11
2 = 11×10
1×2 = 55 pairs of letters that might fall off.
The 55 pairs can be classified into the pair OO, 5 other pairs from the set {H, O, O, N}, 4×7 = 28 pairs
consisting of one letter from {H, O, O, N} and one from {C, A, T, T, A, G, A}, 3 pairs AA and 1 pair TT,
and 17 other pairs from {C, A, T, T, A, G, A}.
If OO fell off, the sign will always read correctly.
If one of 5 other pairs from {H, O, O, N} fell off, the letters will always look right side up, but will be
interchanged with probability 1/2.
For the 28 pairs consisting of one letter from {H, O, O, N} and one from {C, A, T, T, A, G, A}, the
restored sign is correct with probability 1/4, incorrect but readable with probability 1/4, and one letter
appears inverted with probability 1/2.
For the AA and TT pairs, the sign is correct with probability 1/4, has one letter inverted with probability
1/2 and has both letters inverted with probability 1/4.
The remaining 17 pairs are put back correctly with probability 1/8, interchanged but upright with
probability 1/8, have one letter upside down with probability 1/2, and have both letters upside down with
probability 1/4. This gives
(a) P(sign reads CHATTANOOGA) = [ ]
1 + 5×1
2 + 28×1
4 + 4×1
4 + 17–1
8×1
55 = 109
440 ≈ 25%. Not Bad!
(b) P(sign readable but not CHATTANOOGA) = [ ]
5×1
2 + 28×1
4 + 17×1
8×1
55 = 93
440
(c) P(one letter seems to be upside down) = [ ]
28×1
2 + 4×1
2 + 17×1
2×1
55 = 196
440 = 49
110
(d) P(two letters seem to be upside down) = [ ]
4×1
4 + 17×1
4×1
55 = 42
440 = 21
220
Sanity check: The sum of the probabilities is (109+93+196+42)/440 = 1.
(e) All letters seem to be right side up includes the case when the sign reads CHATTANOOGA! Now,
P(at least one vowel|all seem to be right side up) = P(all seem to be right side up∩ at least one vowel)
P(all seem to be right side up)
We know from (a) and (b) that the denominator is (109+93)/440 = 101/220.
Now, our list above can be further classified into
1 pair OO and 2 pairs each of the forms HO and NO (for these the letters always appear to be right side up),
14 of the form O and one from {CATTAGA} and 3 each of the form HA and NA (for these, the probability
that the letters seem right side up is 1/2),
3 of the form AA, and 3 each of the form CA and GA, and 6 of the form TA (for these the probability that
tbe letters seem to be right side up is 1/4). All other pairs {from the set CHTTNG} do not include a vowel.
Check: P(at least one vowel) = 1 – P(no vowel) = 1 – [(6×5)/(1×2)]/55 = 40/55
= (1 + 2 + 2 + 14 + 3 + 3 + 3 + 3 + 3 + 6)/55 so we got 'em all!
From all this, we can readily compute P(all seem to be right side up ∩ at least one vowel)
= [(1+2+2)×1 + (14 + 3 + 3)×1/2 + (3 + 3 + 3 + 6)×(1/4)]/55 = 75/220.
Hence, P(at least one vowel | all seem to be right side up) = 75
220 / 101
220 = 75
101.
(f) The designated driver can correctly identify the letters that fell down if and only if the driver can see that two
letters are obviously switched (possibly being turned upside down) or if two letters both appear to be upside
down but in their correct places. Note that these are disjoint possibilities. For the various pairs of different
types, we see that if OO fell down, the sign will still read CHATTANOOGA and the driver will not be able
to identify with certainty which letters fell down.. If one of the 5 other pairs from {H, O, O, N} fell down,
the driver can identify them if and only if they have been switched (probability 1/2); whether they have been
flipped over doesn’t matter. For the 28 pairs consisting of one letter from {H, O, O, N} and one from
{C, A, T, T, A, G, A}, the two letters must be swapped in order to be identifiable (probability 1/2), while
for the 3 + 1 pairs AA and TT, both letters must be upside down to be identifiable with certainty. Finally,
the remaining 17 other pairs from {C, A, T, T, A, G, A}, either both letters must be in the wrong place
(probability 1/2) or they must be in the right place and both upside down (probability 1/8).
Hence, P{driver can identify correctly} = [ ]
1
55×0 + 5
55×1
2 + 28
55×1
2 + 4
55×1
4 + 17
55×5
8 = 225
440 = 45
88 > 50%!!
2. 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.
