Download Mathematics 4255 Midterm 1 with solutions

Mathematics 4255
Midterm 1 with solutions
February 19, 2013
Partial credit will be awarded for your answers, so it is to your advantage to explain your
reasoning and what theorems you are using when you write your solutions. Please answer
the questions in the space provided and show your computations.
Good luck!
I
II
III
IV
V
VI
Total
Name:
1
I. (10 points) Prove the following statements:
T
1. If P (A) = 0, then what is P (A B)?
T
2. If P (A) = 1, then what is P (A B)?
Hint: Let us recall that B can be rewritten as follows
\ [ \ B= B A
B Ac .
Solutions:
Let us denote by S the sample space.
T
T
T
1. Since A B ⊂ A then 0 ≤ P (A B) ≤ P (A) = 0. Hence P (A B) = 0.
2. P (A) = 1 implies that P (Ac ) = 1 − P (A) = 0. Now,
[
P (B) = P (B S)
[ \
= P (B (A Ac ))
\ [ \
= P (B A) (B Ac )
\ \ = P B A + P B Ac
\ = P B A .
T
We have used the fact that B Ac ⊂ Ac hence
\ \ c
c
0 ≤ P B A ≤ P (A ) = 0 =⇒ P B Ac = 0.
We deduce that
\ P B A = P (B).
2
II. (10 points) Independent flips of a coin that lands on heads with probability p are made.
What is the probability that the first four outcomes are
1. H, H, H, H.
2. T, H, H, H.
Hint: Let us denote by Hi the event that head occurs at the ith flip.
Solutions: P (Hi ) = p. Moreover, the events Hi are independent.
1.
P (H, H, H, H) = P (H1
\
H2
\
H3
\
H4 ) = P (H1 )P (H2 )P (H3 )P (H4 ) = p4 .
2.
P (T, H, H, H) = P (H1c
\
H2
\
H3
\
H4 ) = P (H1c )P (H2 )P (H3 )P (H4 ) = (1 − p)p3 .
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III. (10 points) A total of 48 percent of women (0.48) and 37 percent of the men (0.37)
that took a certain ”quit smoking” class remained nonsmokers for at least one year after
completing the class. These people then attended a success party at the end of the year. If
62 percent (0.62) of the original class was male,
1. what percentage of the original class attended the party?
2. what the percentage of those attending the party were women?
Hint: Denote by A the event that this person attends the party (can be a man or a women).
Denote by W the event that this person is a woman and by W c the event that this person
is a man.
Solutions: Here, we are given the following probabilities: P (W ) = 0.38, P (W c ) = 0.68
P (A|W ) = 0.48, P (A|W c ) = 0.37.
1.
P (A) = P (A|W )P (W ) + P (A|W c )P (W c )
= (0.48)(0.38) + (0.37)(0.62) ≈ 0.4118
2.
T
P (W A)
P (W |A) =
P (A)
P (A|W )P (W )
=
P (A|W )P (W ) + P (A|W c )P (W c )
(0.48)(0.38)
≈
0.4118
≈ 0.443.
4
VI. 10 points Let X denote a random variable that takes on the values -1, 0, and 1 with
respective probabilities P (X = −1) = 0.2 P (X = 0) = 0.5 P (X = 1) = 0.3. Let Y be
another random variable defined by Y = X 2 .
1. Compute the probability density (mass) function of Y .
2. Compute the expected value of Y .
3. Compute the variance of Y .
Solutions: Since X takes values in {−1, 0, 1} then Y = X 2 takes values in {0, 1}.
1. Let us compute the probability mass of Y .
P (Y = 0) = P (X 2 = 0) = P (X = 0) = 0.5
P (Y = 1) = P (X 2 = 1)
= P (X = −1 or X = 1)
= P (X = −1) + P (X = 1) = 0.2 + 0.3 = 0.5.
Hence, pY (1) = 0.5 and pY (0) = 0.5
2.
E(Y ) =
X
i pY (i) = 0 × pY (0) + 1 × pY (1) = 0.5.
i
E(Y 2 ) =
X
i2 pY (i) = 0 × pY (0) + 1 × pY (1) = 0.5.
i
V ar(Y ) = E(Y 2 ) − (E(Y ))2
= 0.5 − (0.5)2 = (0.5) × (0.5) = 0.25
5
V. (10 points) Prove that if E1 , E2 , . . . , En are independent events, then
n
[ [ [ Y
P E1 E2 · · · En = 1 −
(1 − P (Ei )) .
i=1
Solutions: Since E1 , E2 , . . . , En are independent events, then E1c , E2c , . . . , Enc are also independent events.
Using the Di Morgan’s law and the independence of events, we get that
[ [ [ [ [ [ c
P E1 E2 · · · En = 1 − P E1 E2 · · · En
\ \ \ = 1 − P E1c E2c · · · Enc
= 1 − P (E1c ) × P (E2c ) × . . . P (Enc )
= 1 − (1 − P (E1 )) × (1 − P (E2 )) × . . . (1 − P (En ))
n
Y
=1−
(1 − P (Ei ))
i=1
6
VI. (Bonus 10 points) Independent trials, each resulting in a success with probability p
or failure with probability q = 1 − p, are performed. Find the probability that a run of n
consecutive successes occurs before a run of m consecutive failures.
Hint: Denote by E the event that a run of n consecutive successes occurs before a run
of m failures. And denote by H the event of a success.
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