University Problem Set #4: Solutions ECE 313
of Illinois Page 1 of 2 Spring 2003
1. Y = 1, 2, 3 passengers are left behind according as X = 6, 7, or 8. Since X takes on values 6, 7, 8 with
with probabilities 28
256, 8
256, 1
256 respectively, we readily find that E[Y] = 1×28 + 2×8 + 3×1
256 = 47
256.
2.(a), (b)
0
0.2
0.4
0 1 2 3 4 5 6 7 8 9 10
p = 0.1
p = 0.9 0
0.1
0.2
0.3
012345678910
p = 0.25
p = 0.75
0
0.1
0.2
0.3
0 1 2 3 4 5 6 7 8 9 10
p = 0.4
p = 0.6
p = 0.5
0
0.2
0.4
012345678910
(c) From the table, the probabilities for any given value of p and 1–p are just “reverses” of each other in the
sense that P{X = k} for probability p is just P{X = 10–k} for probability 1–p.
(d) For p = 0.1, 0.25. 0.4, 0.5, 0.6, 0.75, and 0.9, P{X = k} is largest for k = 1, 2, 4, 5, 6, 8, and 9
respectively exactly as predicted by the theory (would I lie to you?)
(e) The mean is np and the mode is (n+1)p , so the difference between the two values is ≤ p ≤ 1. Why?
3.(a) P(same on all three days) = (0.2)3 + (0.5)3 + (0.3)3 = 0.16.
(b) P(same on two of three days) = 3 ×{(0.2)2 × [1 – 0.2] + (0.3)2 × [1 – 0.3] + (0.5)2 × [1 – 0.5]} = 0.66.
(c) P(different on all three days) = 3![0.2 × 0.5 × 0.3] = 0.18. Alternatively, we can compute this as
1 – 0.16 – 0.66. (Why?)
4.(a) X takes on values –6, 6, 12, 18.
(b) If $6 is bet on i, then you lose it if all three dice show one of the 5 non-i numbers.
Hence P{X = –6} = 53/63 = 125
216. On the other hand, you win $6 if one of the three dice shows i and the
other two have non-i numbers. Hence, P{X = 6} = 3•(1•52)/63 = 75
216. By a similar argument, P{X = 12}
is 3•(12•5)/63 = 15
216, and P{X = 18) = 13/63 = 1
216. Sanity check: 125 + 75 + 15 + 1 = 216, so we have
not left anything out.
(c) E[X] = ∑
u•p(u) = 125•(–6) + 75•6 + 15•12 + 1•18
216 = – 102
216 = –17
36 i.e. a loss of roughly 47¢ per game.
(d) The chance of the three dice showing three different numbers is 6•5•4
216 = 5
9. In this case, you come out even
since you win $3 on the three numbers showing, but lose $3 on the three no-shows. The chance that the
three dice show the same number is 6• 1
216 = 1
36 in which case you win $3 on the winning number but lose
$5 on the 5 no-shows for a net loss of $2. The probability that exactly two numbers are identical is thus
15
36 = 5
12 in which case, you win $2 on the pair and $1 on the singleton, but lose $4 on the other no-shows
for a net loss of –1. Thus, Y is a random variable taking on values 0, –1, –2, with probabilities as found
above. (Remind me once again why you are bothering to play this game at all?), and its expected value is
