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1. E, F, and G are events defined on a sample space
(a) Only E occurs: EFcG c. Note that the expression E means that E occurs without excluding F or G; they
may or may not have occurred. (b) Both E and G but not F: EF cG. (c) At least one of the three occur. E ∪ F ∪ G. The more cumbersome expression E ∪ F ∪ G ∪ EF ∪ EG ∪ FG ∪ EFG is equivalent to the simpler one (Prove it!)
(d) At least two occur. EF ∪ EG ∪ FG. (e) All three occur. EFG. (f) None occur. EcF cG c
(g) At most one occurs. EcF c^ ∪ EcG c^ ∪ F cG c^ (Compare (d). The given expression is just the one describing
the event that at least two of the events Ec, F c, G c^ occurred (and hence at most one of E,F, G occurred!))
(h) At most two occur. Ec^ ∪ F c^ ∪ G c^ (Compare (c). This is just the event that at least one of the events Ec,
F c, G c^ occurred (i.e. at most two of E, F, G occurred). Alternatively, (EFG)c^ since all three do not occur.
(i) Exactly two occur. EFGc^ ∪ EF cG ∪ EcFG or (EF ∪ EG ∪ FG) – EFG. (j) At most three occur. Ω
2.(a) The binomial theorem states that (1 + x)n^ = ∑
i=
n
n
i x^
i (^) where
n
i , the coefficient of x^
i (^) , is the number of
events that contain i outcomes. Putting x = 1 gives 2n^ = ∑
i=
n
n
i , i.e. the total number of events is 2
n (^).
Putting x = –1 and noting that (–1) i^ is +1 or –1 according as i is even or odd, we get that
i even
n
i –^ ∑
i odd
n
i which in turn implies that^ ∑
i even
n
i =^ ∑
i odd
n
i = 2^
n–1 (^) (Why is each sum 2n–1 (^) ?)
Alternatively, suppose that n is odd and consider an event A of odd cardinality k. Then, Ac^ has even cardinality n–k. Thus, corresponding to every event of odd cardinality, there is an event of even cardinality (and vice versa). It follows that there there must be equal numbers of each type of event. Unfortunately, this proof does not work for even n, but the following always works! Let x denote one of the outcomes in Ω, and consider the 2 n–1^ events (subsets of Ω) that do not include x. Thus, ø is included among these events but Ω is not. Suppose that i such events are of odd cardinality and j of even (where i + j = 2 n–1^ ). The remaining 2 n–1^ events all contain x, and can be thought of as being formed by including x in each of the 2n–1^ events previously considered, e.g. changing {a,b,c,d} to {a,b,c,d,x} and ø to {x} and {a,b,c,d,…} to {a,b,c,d,…,x} = Ω. But, including x in an event changes the cardinality from odd to even and vice versa. Thus, i events that include x have even cardinality and j have odd cardinality. Hence, i+j = 2n–1^ events are of even cardinality and i+j = 2n–1^ events are of odd cardinality.
(b) Pair up each event A with its complement Ac. As a special case, Ω is paired with ø, and note that we thus
have matched up the 2 n^ events into 2n–1^ pairs. Now, consider the sum of the probabilities of all 2n^ events. Each pair contributes a total of P(A) + P(A c) = 1 to the sum. Since there are 2n–1^ such pairs, the sum of all the probabilities is 2n–1^. Dividing by 2 n^ , we find that the “average probability” is 1/2. Note that we have not needed the assumption that all the n outcomes are equally likely — any assignment of probabilities p 1 ,
p 2 , … pn to the n outcomes (where each pi ≥ 0 and ∑
i=
n pi = 1) gives the same result that the “average
probability” is 1/2; because it is always true that P(A) + P(Ac) = 1. A longer way of obtaining the same result (with equally likely outcomes) is to note that an event containing i outcomes has probability i/n so that we are finding the sum
i=
n
n
i ×^
i
n =^ ∑
i=
n
n
i ×^
i n =^ i
n n n n i
i × ×…×
× − ×…× − +
1
∑ ×^
i n =
1 × ×…× −
− ×…× − +
∑ = i
n n n i
i
j=
n-
n-1
j = 2^
n–
Or, differentiate both sides in the statement of the binomial theorem and then set x = 1 to get n(1 + x)n-1^ =
i=
n
n
i ix^
i–1 ⇒ n2 n–1 = ∑
i=
n
n
i ×^ i =^ ∑
i=
n
n
i ×^ i. This can also be interpreted to mean that adding up the
cardinalities of all the subsets of Ω gives a total count of |Ω|× 2 |Ω|–1^ (There are n× 2 n–1^ ones in a truth-table!)
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(c) If n is even, then ^
n n/2 events have probability 1/2. If n is odd, no event has probability 1/2.
(d) Let x denote the outcome that has occurred. Then, all the 2 n–1^ events that have occurred can be expressed
as {x} ∪ A where A denotes a subset of the remaining n–1 outcomes. The sum of the cardinalities of all these 2 n-1^ sets is 2n-1^ + (n-1)× 2 n-2^ where the first term comes from the x in each set and the second from the 2 n-1^ subsets A. Divide by n to get the total probability and by 2n-1^ to get that the average probability of the 2 n-1^ events that occurred is (n+1)/(2n) = (1/2) + 1/(2n)which is just slightly larger than 1/2.
3.(a) From DeMorgan's theorem and P(AcBc) = 0.3, we get that P(A∪B) = 0.7.
P(AB) = P(A) + P(B) – P(A∪B) = 0.1. P(ABc) = P(A) – P(AB) = 0.35.
(b) P(ABCc) = P(AB) – P(ABC) = 0.098. Note that C can be partitioned into ABC, (A⊕B)C and AcBcC,
where (A⊕B)C is the shaded set in the figure shown. Since P(AcBcC) = 0, we get that P(C) = P(ABC) + P(A⊕B)C) = 0.006.
(c) P(AcBCc) and P(A∪C) cannot be computed.
A
C
B
Figure for Problem 3
0.
(^0) 0.
0.3^ 0.
0.35 0.
4.(a) A = (ddddddd1: d = don't care if it is 0 or 1} giving |A| = 2 7 = 128; P(A) = 2 7 /2 8 = 1/2.
(b) The shift register contains 4 1’s which can be chosen from the 8 positions in
4 =^1234
× × ×
× × ×
= 70 ways
(the remaining 4 positions have 0’s in them.) Hence, B contains 70 outcomes and P(B) = 70/2^8 = 35/128. (c) AB is the set of shift register contents with 4 1’s, one of which is in the least insignificant bit posiiton. The
remaining 3 1’s can be in any of
3 =^123
× ×
× ×
= 35 positions. Hence, P(A∩B) = 35/2 8 ;
P(A∪B) = P(A) + P(B) – P(A∩B) = (128+70–35)/2 8 = 163/2 8 ; P(A ⊕ B) = P(A) + P(B) – 2P(A∩ B) = (128+70–70)/2 8 = 1/2. 5. (a) The outcomes are all the vectors of length 5 in which 3 entries are b and 2 are c. The days to serve
broccoli can be chosen in
3 = 10 ways, so^ |Ω|^ = 10. (b) Mom chooses broccoli with probability 3/5 from the five choices she has on Monday. Alternatively,
broccoli is served on two other days besides Monday, and these days can be chosen in
2 = 6 ways. Hence, P(broccoli on Monday) = 6/10 = 3/5. (c) The third broccolous day can be any of 3, and hence probability is 3/10. (d) 1/10. 6. We use the Venn diagram/Karnaugh map shown below.
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(b) Let x = ln 2. Then, P(Ω) = ∑
n=
∞ (ln 2) n 2(n!) = 2^
n=
∞ x n n! = 2^
–1 (^) • exp(x) = 1.
P(outcome is an even number) = ∑
n=
∞ (ln 2) 2n 2(2n!) = 2^
n=
∞ x 2n (2n!) = 2^
–1 (^) • cosh(x) =^5 8 since x = ln 2.
9.(a) The number of weeks that your investment doubles in value is a binomial random variable Y with parameters (5,1/2). Since the investment halves in value during the remaining 5– Y weeks, and each halving cancels one doubling, we have that X = 32•2 2 Y –5^. The possible values of X are 1, 4, 16, 64, 256, and 1024, corresponding to Y = 0, 1, 2, 3, 4, 5 respectively. (b) P{ X = 1} = P{ Y = 0} = 1/32. P{ X = 4} = P{ Y = 1} = 5/32. P{ X = 16} = P{ Y = 2} = 10/32. P{ X = 64} = P{ Y = 3} = 10/32. P{ X = 256} = P{ Y = 4} = 5/32. P{ X = 1024} = P{ Y = 5} = 1/32. (c) E[ X ] = 1•(1/32) + 4•(5/32) + 16•(10/32) + 64•(10/32) + 256•(5/32) + 1024•(1/32) = 97.65625. The TV commercial understates the performance — undoubtedly a first! (d) P{ X < 32} = P{ X = 1} + P{ X = 4} + P{ X = 16} = 1/2. (e) There is a 50% chance of losing money on this investment. Most people do not mind making investments if the amount at risk is small but the payoff from a win is enormous, e.g. lottery tickets are a losing bet for most buyers, but people don’t mind an almost sure loss of a dollar because all they see is the huge payoff. Matters are considerably different when the amounts at risk are large, and most people tend to choose more conservatively. This explains why the local 7-11 sells lottery tickets but Neiman-Marcus does not. 10.(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} = 5 3 /6 3 =
- 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•5 2 )/6^3 =
- By a similar argument, P{ X^ = 12}
is 3•(1 2 •5)/6 3 =
216 , and P{ X^ = 18) = 1^
- Sanity check: 125 + 75 + 15 + 1 = 216, so we have not left anything out.
(c) E[ X ] = ∑u•p(u) =
36 i.e. a loss of roughly 47¢ per game.
(d) The chance of the three dice showing three different numbers is
- 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•
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
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 E[ Y ] =
36 just as before. Splitting your bet six ways has no effect on your losses!
