STAT 205 Probability Theory Midterm
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Probability Theory
Midterm
1. (5pts+5pts=10pts)
A bag contains 8 pairs of shoes; each pair is a different style. You pick two random shoes from the bag.
(a) What is the probability that the two shoes you picked out are a pair; i.e. left and right of the same style?
Solution
16*1/(16*15)=1/15
(b) What is the probability you picked one left shoe and one right shoe?
Solution
16*8/(16*15)=8/15
2. (5pts+5pts=10pts)
A cooler contains 15 cans of Coke and 10 cans of Pepsi. Three are drawn from the cooler without replacement.
(a) What is the probability that all three are the same brand?
(b) What is the conditional probability that all three are Pepsis given that they are all the same brand?
Solution(a)
(b)
3.(4pts+4pts+4pts=12pts)
(a) Let 
and 
be two events, 0 < 
< 1 . Show that if 
(
) ≥ 1 − 
and
(
) ≥ 1 − 
, then
(
) ≥ 1 − 2
.
Proof Noting that 
(
∪ 
) ≤ 1,
(
) = 
(
) + 
(
) − 
(
∪ 
) ≥ (1 − 
) + (1 + 
) − 1 = 1 − 2
.
(b) Prove that if A and B are mutually exclusive, then
(
|
∪ 
) = 
Proof
Since A and B are mutually exclusive, 
(
∪ 
) = 
(
) + 
(
),
therefore,
(
|
∪ 
) = 
(
∩ (
∪ 
)) = 
(
)
(c) Let 
be a exponential random variable with parameter 
, λ > 0, i.e., 
(
) = 
−
, 
> 0.
Show that 
(
> 
+ 
|
> 
) = 
(
> 
) for 
> 0, 
> 0 (the “memoryless property”).
Proof
First, let’s compute the cdf of 
,
(
) = 
(
≤ 
) = ∫ 
−
= [1 − 
−
].
Hence,
(
> 
+ 
|
> 
) = =
4. (5pts+5pts=10pts)
In a certain country 
of the population are college graduates. 
of the college
graduates wear glasses, but only 
of the non-college graduates wear glasses.
(a) Find the probability that a randomly chosen person wears glasses.
(b) You see a random person in the street wearing glasses, what is the probability that she or he is a college graduate?
Solution
(a)
(b)
The numerator
5. (5pts)
Suppose that, for a discrete random variable 
, 
(
) = 3 and 
(
(
− 4)) = 6.
Find the variance of 2
+ 3.
Solution
E(X2 − 4X) = 6 implies that 
(
2) − 4
(
) = 6. Substituting 
(
) = 3 in this relation gives 
(
2) = 18. Hence,
(
) = 
(
2) − (
(
))2 = 18 − 32 = 9,
and
(2
+ 3) = (2)2 × 
(
) = 4 ∗ 9 = 36.
6. (5pts+5pts=10pts)
Suppose 
is a Poisson random variable with parameter λ, and
(
= 1) = 
(
= 2).
(a) Find 
.
(b) Compute the expected value of 
.
Solution
(a) Since 
(
= 1) = 
(
= 2), we have
1 
−
= 
2 
−
1! 2! .
Solving the preceding equation for λ gives
λ = 2.
(b)
= 
7. (5pts+5pts=10pts)
Suppose 
is uniform on the interval from 1 to 2, let 
= 1/
.
(a) Compute the pdf of 
.
(b) Compute the expected value of 
.
(b)
Alternatively,
8. (5pts+5pts+5pts+5pts=20pts)
A continuous random variable X has cumulative distribution function
Solve the following. (a) Determine the constants
(
) = {
2,
if 
≤ 0
if 0 < 
< 1
if 
≥ 0
(b) Find the probability distribution function of 
. Make sure that you provide a formula for 
(
) that is valid for all 
.
(c) Calculate the expected value of 
.
(d) Calculate the variance of 
.
9. (5pts)
A certain population is comprised of half men and half women. Consider a random sample of 100,000 persons. Write down the probability that the percentage of men in the sample is somewhere between 49% and 51% as an approximation to the standard Gaussian and compute your answer. You will need to use 
(2) ≈ 0.9772.
Solution
Let 
denote the number of men in the sample.
10. (5pts+3pts=8pts)
The random variable X is called double exponentially distributed if its density function is given by
(
) = 
−|
|, −∞ < 
< ∞ .
(a) Find the value of 
.
Solution
We must have 
−|
|
= 1, thus
c = 
= 
= 
.
(b) Prove that 
(
2
) = (2
)! and 
(
2
+1) = 0.
Proof
(
2
+1) = 
2
+1
−|
|
= 0, because the integrand is an odd function. 
(
2
) = 
2
−|
|
= 
2
−
, because the integrand is an even function. We now use induction to prove that 
−
= 
!. For 
= 1, the
integral is the expected value of an exponential random variable with parameter 1; so it equals to 1 = 1! . Assume that the identity is valid for 
−
1. Using integration by parts, we show it for 
.
∞
−
= −[−
−
]0(∞) + 
∞
−1
−
= 0 + 
(
− 1)! = 
!.
Hence 
(
2
) = (2
)!.
2022-07-12