Download Problems and Solutions for ECE 313: Calculating Probabilities of Mutually Exclusive Events and more Assignments Statistics in PDF only on Docsity!
of Illinois Page 1 of 4 Spring 2002
Assigned: Wednesday, January 30
Due: Wednesday, February 6
Reading: Ross, Chapters 2.1–2.5, 4.1, 4.3–4.5, 4.
Noncredit Exercises: (Do not turn these in) Chapter 4, Problems: 2, 7, 13, 28, 35, 39, 40-43;
Theoretical Exercises: 11, 13, 15.
Problems:
1. Let A, B, and C denote three events defined on a sample space Ω, and suppose that
P(A) = 0.3, P(B) = 0.3, P(C) = 0.5, and P(A ∩ Bc) = P(Ac
∩ Bc
∩ C) = 0.2.
Find P(A ∩ B), P(Ac
∩ B), P((A ∪ B ∪ C)c), and P( B ∩ Cc).
If any of these probabilities cannot be computed from the given information, find upper and
lower bounds (preferably other than 1 and 0 !!) on these probabilities.
1. (a) A Karnaugh map/Venn diagram is very useful in solving problems such as these.
A
B
C
A
B
Since A is the union of the disjoint sets A ∩ B and A ∩ B
c ,
P(A) = P(A ∩ B) + P(A ∩ B
c ) ⇒ P(A ∩ B) = P(A) – P(A ∩ B
c ) = 0.3 – 0.2 = 0.1.
Similarly, P(A
c ∩ B) = P(B) – P(A ∩ B) = P(B) – P(A) + P(A ∩ B
c ) = 0.3 – 0.3 + 0.2 = 0.2.
A
B
0.2 C
A
B
Next, P(A ∪ B ∪ C) = P(A ∪ B ) + P((A ∪ B )
c ∩ C) = P(A ∪ B ) + P(A
c ∩ B
c ∩ C) (why?)
= P(A) + P(B) – P(A ∩ B) + P(A
c ∩ B
c ∩ C) = 0.3 + 0.3 – 0.1 + 0.2 = 0.7,
and, of course, P((A ∪ B ∪ C)
c ) = 1 – P(A ∪ B ∪ C) = 1 – 0.7 = 0.3.
P(B ∩ C
c ) cannot be determined, but we can be sure that P(B ∩ C
c ) ≤ P(B) = 0.3. On the other hand, the
smallest possible value of P(B ∩ C
c ) is 0 when P(A ∩ B
c ∩ C) = 0 and P(A ∩ B
c ∩ C
c ) = 0.2.
Note that the value of P(C) is not used anywhere except in deciding that P(B ∩ C
c ) cannot be determined
and in determining the bounds on P(B ∩ C
c ).
2. The experiment consists of picking a student from the set of all UIUC students registered
this semester. It is not necessary to assume that all students are equally likely to be picked,
but you may make this assumption if it makes you feel happier and more confident.
of Illinois Page 2 of 4 Spring 2002
(a) Let A and B denote the events that the student picked has had respectively four years of
science (FYS) and calculus in high school. Let P(A) = 0.45 and P(B) = 0.35. If the
probability that the student had neither FYS nor calculus is 0.3, what is the probability that
the student had both FYS and calculus? What is the probability that the student had FYS
but not calculus?
(b) Let C denote the event that the student is registered in ECE 313, and let A and B be as in
part (a). Suppose that P(A∩B∩C)=0.002. What is the probability that the student picked
is not registered in ECE 313, but did have both FYS and calculus? If the probability that
the student picked is registered in ECE 313, and has had either FYS or calculus (but not
both) is 0.004, and if students who had neither FYS nor calculus did not register in ECE
313, what is P(C)?
(c) Using the data given in parts (a) and (b), which of the following probabilities can you
compute? It is not necessary to actually compute each probability.
P(A ∪ C), P(A∪B∪C), P(A ∪ B∪ Cc), P(AcBcCc), P(AcBCc), P(ABCc)
2.(a) From DeMorgan's theorem and P(A
c B
c ) = 0.3, we get that P(A∪B) = 0.7.
P(AB) = P(A) + P(B) – P(A∪B) = 0.1. P(AB
c ) = P(A) – P(AB) = 0.35.
(b) P(ABC
c ) = P(AB) – P(ABC) = 0.098. Note that C can be partitioned into ABC, (A⊕B)C and A
c B
c C,
where (A⊕B)C is the shaded set in the figure shown below.
Since P(A
c B
c C) = 0, we get that P(C) = P(ABC) + P(A⊕B)C) = 0.006.
(c) P(A
c BC
c ) and P(A∪C) cannot be computed.
A
B
C
3. Let A, B, C denote three events defined on a sample space. Show that
P(A) + P(B) + P(C)
≤ P(A ∪ B ∪ C) ≤ P(A) + P(B) + P(C)
3. Since A is a subset of A ∪ B ∪ C, we know that P(A) ≤ P(A ∪ B ∪ C). Similar inequalities hold for B
and C as well. Adding the three inequalities gives that P(A) + P(B) + P(C) ≤ 3P(A ∪ B ∪ C), and thus it
follows that [P(A) + P(B) + P(C)]/3 ≤ P(A ∪ B ∪ C).
To prove the other inequality, note that P(E ∪ F) = P(E) + P(F) - P(EF) for any events E and F, and hence
P(A ∪ B) ≤ P(A) + P(B) holds for events A and B (with equality if P(A∩B) = 0.) 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).
4. Find P(A ∪ (B
c
∪ C
c
c
) in each of the following four cases:
(a) A, B, and C are mutually exclusive events and P(A) = 1/3.
(b) P(A) = 2P(BC) = 4P(ABC) = 1/2.
(c) P(A) = 1/2, P(BC) = 1/3, and P(AC) = 0.
(d) P(A
c
∩ (B
c
∪ C
c
4. P(A ∪ (B
c ∪ C
c )
c ) = P(A∪(B∩C)) by DeMorgan's theorem.
(a) P(B∩C) = 0, and therefore P(A∪(B∩C)) = P(A∪ø) = P(A) = 1/3.
(b) P(A∪(B∩C)) = P(A) + P(B∩C) - P(A∩B∩C) = 1/2 + 1/4 - 1/8 = 5/8.
(c) P(A∪(B∩C)) = P(A) + P(B∩C) - P(A∩B∩C) = 1/2 + 1/3 - 0 = 5/6. (Why?)
(d) (A
c ∩ (B
c ∪ C
c ))
c = A ∪ (B
c ∪ C
c )
c by DeMorgan's theorem. Hence,
P(A ∪ (B
c ∪ C
c )
c ) = 1 - P(A
c ∩(B
c ∪ C
c )) = 1 - 0.6 = 0.4.
of Illinois Page 4 of 4 Spring 2002
culinary skills, makes independent decisions (e.g. without taking into account your opinion
that Cheetos is a vegetable suitable for serving with any entree) about the vegetable to serve
each day. Over a three day period, what is the probability that
(a) she serves the same vegetable on all three days?
(b) she serves the same vegetable exactly two days out of three?
(c) she serves different vegetables on the three days?
7.(a) P(same on all three days) = (0.2)
3
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.
