Download Solutions to Problem Set 2 in ECE 413 at University of Illinois Spring 2008 - Prof. Dilip and more Assignments Statistics in PDF only on Docsity!
University of Illinois Spring 2008
ECE 413: Solutions to Problem Set 2
- Define events C = {At least one of O and G occurs} and D = {At least one of O and G does not
occur}. The Karnaugh map below shows the various events of interest:
We are given that P (C) = P (O ∪ G) = 0.65 and P (D) = P (O
c ∪ G
c ) = 0.7. Now, note that C ∪ D = Ω while C ∩ D = (O ∪ G) ∩ (Oc^ ∪ Gc) = (O ∩ Gc) ∪ (Oc^ ∩ G) = O ⊕ G. Consequently,
P (C ∪ D) = P (Ω) = 1 = P (C) + P (D) − P (O ⊕ G) = 0.65 + 0. 7 − P (O ⊕ G) from which we get that
P (O ⊕ G) = 0.35.
- (1 + x)
n
n
0
n
1
x +
n
2
x
2
n
3
x
3
n
4
x
4
(1 − x)
n
n
0
n
1
x +
n
2
x
2 −
n
3
x
3
n
4
x
4 − · · ·
from which we get that (1 + x)
n
n = 2
[(
n
0
n
2
x
2
n
4
x
4
]
. Set x = 1 and note
that the left side has value 2n^ while the right side is twice the number of sets with an even number of
elements. More simply, set x = 1 in (1 − x)
n to get
n
0
n
1
n
2
n
3
n
0
n
2
n
4
n
1
n
3
n− 1 .
- (a) For any event A, P (A) ≤ 1. Hence, P (A ∪ B) = P (A) + P (B) − P (A ∩ B) ≤ 1. It follows that
P (A ∩ B) ≥ P (A) + P (B) − 1
(b) i. Since A is a subset of A ∪ B, we know that P (A) ≤ P (A ∪ B). Similarly P (B) ≤ P (A ∪ B).
Adding the two inequalities gives that P (A) + P (B) ≤ 2 P (A ∪ B), and thus it follows that
P (A) + P (B)
≤ P (A ∪ B)
with equality iff P (A) = P (B) = P (A ∪ B) or equivalently, P (A) = P (B) = P (A ∩ B)
To prove the other inequality, note that P (A ∪ B) = P (A) + P (B) − P (A ∩ B) for any events
A and B. Since P (A ∩ B) ≥ 0, it follows that
P (A ∪ B) ≤ P (A) + P (B)
with equality iff P (A ∩ B) = 0. This bound is sometimes referred to as the union bound.
ii. As shown in the previous part, P (A) ≤ P (A ∪ B ∪ C), P (B) ≤ P (A ∪ B ∪ C) and P (C) ≤
P (A∪ B ∪C). Adding the three inequalities gives that P (A)+ P (B)+P (C) ≤ 3 P (A∪ B ∪C), and thus it follows that
P (A) + P (B) + P (C)
3
≤ P (A ∪ B ∪ C)
with equality iff P (A) = P (B) = P (C) = P (A ∪ B ∪ C). In the previous part, we proved that P (A ∪ B) ≤ P (A) + P (B) holds for events A and B.
Let D denote A ∪ B. Then,
P (A ∪ B ∪ C) = P (D ∪ C) ≤ P (D) + P (C) = P (A ∪ B) + P (C) ≤ P (A) + P (B) + P (C).
Equality holds iff P (A∩B) = P (A∩C) = P (B∩C) = 0 (which implies that P (A∩B∩C) = 0.)
- (a) Each club must have at least one member (i.e. the leader) who can be chosen to be any of the
n FOMDLIUans. The remaining members of the club can be any subset of the remaining n − 1
FOMDLIUans. Since there are 2
n− 1 such subsets, we conclude that the number of clubs is n 2
n− 1 .
More laboriously, we have n = n
(n− 1
0
possibilities for clubs with 1 member, n
(n− 1
1
possibilities
for clubs with 2 members, n
n− 1 2
possibilities for clubs with three members, and so on. Putting
this together, we get:
Number of clubs = n
n − 1
0
n − 1
1
n − 1
2
n − 1
n − 2
n − 1
n − 1
= n
[(
n − 1
0
n − 1
1
n − 1
2
n − 1
n − 2
n − 1
n − 1
)]
= n 2
n− 1 .
(b) Using the previous part, we see that the number of clubs with exactly k members is given by
n
(n− 1
k− 1
which is the same as k
(n
k
, as was proved in class. The first displayed sum in 4(a) is
thus
∑n k=1 k
n k
which we counted to be n 2 n−^1 in part (a). More explicitly, we count the number
of clubs by first selecting the members and then choosing the leader from among the members.
Fix a number of members k for a particular club. From a pool of n individuals, there are
(n
k
possibilities. Once we select a club of size k, there are now k possibilities to select a leader. Therefore the total number of clubs is given by:
( n
1
n
2
n
n
· n =
∑^ n
k=
k
n
k
= n 2
n− 1 .
(c)
d
dx
[(1 + x)
n ] = n(1 + x)
n− 1 (by the Chain Rule)
(1 + x)
n
n
0
n
1
x +
n
2
x
2
n
3
x
3
d
dx
[(1 + x)
n ] =
n
1
n
2
x + 3
n
3
x
2
n
4
x
3
Evaluate the derivative of (1 + x)n^ at x = 1 in two different ways and equate the results to get
n 2
n− 1
∑^ n
k=
k
n
k
- (a) The Karnaugh map is as shown in the left hand figure below, with some probabilities marked on
it. Note that the shaded region is the event (A ∩ B) ∪ (B ∩ C) ∪ (A ∩ C).
0.05 0.
︸ ︷︷ ︸
︷ ︸︸ ︷
Bc^ B B
c (^) B
︸ ︷︷ ︸ ︸ ︷︷ ︸ ︸^ ︷︷^ ︸
A
c A
c A A
c A
c A
C
Cc^ Cc
C
︷ ︸︸ ︷
(b) Since A ∩ B is the disjoint union of A ∩ B ∩ C and A ∩ B ∩ C
c , we get that P (A ∩ B) = 0.1 = P (A ∩ B ∩ C) + P (A ∩ B ∩ Cc) = 0.05 + P (A ∩ B ∩ Cc) giving that P (A ∩ B ∩ Cc) = 0.05.
Since P (AB ∪ BC ∪ AC) = 0.3 = P (AB) + P (ABcC) + P (AcBC) while P (AC) = P (ABC) +
P (AB
c C) = 2P (BC) = P (ABC) + P (A
c BC), we readily obtain that P (AB
c C) = 0.15 and
P (A
c BC) = 0.05. Since P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.6 we have that P (A
c B
c ) = 1 − P (A ∪ B) = 0.4. Since P (AcBcC) = P (C) − P (BC) − P (ABcC) = 0.05, we get that
P (cereal snaps, crackles, and pops) = P (A
c ∩ B
c ∩ C
c ) = 0.55.
