EECS 501 SOLUTIONS TO PROBLEM SET #1 Fall 2001
1. Let bi, i = 1 . . . 6 denote the ith ball in the urn, and
bibjdenote that the ith THEN jth balls are taken (note that order matters).
1a. without replacement: Ω has (6)(6 −1) = 30 elements, which are:
Ω = {b1b2, b1b3, b1b4, b1b5, b1b6, b2b1, b2b3, b2b4, b2b5, b2b6. . . b5b6}.
1b. with replacement: Ω has (6)(6) = 36 elements, which are:
Ω = {b1b1, b1b2, b1b3, b1b4, b1b5, b1b6, b2b1, b2b2, b2b3, b2b4. . . b6b6}.
2. Let H=height of man in inches and W=height of woman in inches.
2a. Ω = {(H, W ) : H > 0and W > 0}= (R+)2.(b): E={(H , W ) : 0 < H < W }.
3a. A∪ ∅ =Aand A∩ ∅ =∅ → P r [A] = P r[A∪ ∅] = P r[A] + P r[∅]→P r[∅] = 0.
3b. (EF )∪(E F 0) = Eand (EF )∩(E F 0) = ∅ → P r[E] = P r[EF ] + P r [EF 0].
3c. E∪E0= Ω and E∩E0=∅ → P r[E] + P r [E0] = P r[Ω] = 1 →P r[E] = 1 −P r [E0].
4. (EF 0)∪(F E 0) = E⊕Fand (EF 0)∩(F E 0) = ∅ → P r[E⊕F] = P r[EF 0] + P r [F E0].
5. #3b→P r[E F 0] = P r[E]−P r[EF ] and P r [F E0] = P r[F]−P r[F E ].
#4 →P r[E⊕F] = P r[E] + P r [F]−2P r[EF ] QED. Note P r[E F ] = P r[F E].
6. See overleaf.
7. Let B=∩∞
n=1An=∩∞
n=1(0,1/n). x ∈B⇔x∈An⇔0< x < 1/n for all n.
But for any x > 0,∃N s.t. 1/N < x →x /∈AN→x /∈B. Hence B=∅.
8. Let Z={integers}and Pn={polynomials of degree nwith integer coefficients}.
Pnis 1-1 with Zn+1 since a0+a1z+...+anzn∈Pn↔(a0, a1. . . an)∈ Z n+1 .
A={algebraic irrationals}1-1 with a subset of ∪∞
n=1Pn, since every
A∈ A is the root of a polynomial with integer coefficients (some duplication).
∪∞
n=1Pn=countable union of countable sets=countable→ A at most countable.
{21/n, n = 2,3. . .}⊂A→Aat least countably infinite → A is countably infinite.
