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Department of EECS - University of California at Berkeley EECS 126 - Probability and Random Processes - Spring 2007 Midterm 1: 2/19/
Solutions
You are given a 1-meter long wood stick. You choose two points A and B uniformly and inde- pendently on the stick. You cut the stick at A and at B. You are left with three pieces. What is the probability that you can form a triangle with the three pieces?
Consider the case A < B. The three pieces of stick have lengths X = A, Y = B − A, Z = 1 − B. You can make a triangle if each length is larger than the sum of the other two. That is, we need X > Y + Z, Y > X + Z, Z > X + Y. These inequalities in terms of A, B, C give B ≥ 1 / 2 , A ≤ 1 / 2 , B − A ≤ 1 /2.
The case B < A is symmetric and gives A ≥ 1 / 2 , B ≤ 1 / 2 , A − B ≤ 1 /2.
Combining these two events, we see that they correspond to the sets of pairs (A, B) shown in the figure below:
Figure 1: The shaded triangles correspond to the desired event.
The probability that (A, B) falls in the shaded set is 1/4.
Let X be Poisson with parameter λ > 0. For any positive integer k, calculate
E(X(X − 1)(X − 2) × · · · × (X − k)).
We find
E(X(X − 1)(X − 2) × · · · × (X − k)) =
∑
n≥ 0
n(n − 1)(n − 2) × · · · × (n − k) λn n!
e−λ
= λk+^
∑
n≥k+
λn−k−^1 (n − k − 1)! e−λ^ = λk+^
∑
m≥ 0
λm m! e−λ
= λk+1.
You do not feel too well and you wonder why. The prior probability that you have the flu, some food poisoning, or some other disease D is 10%, 5%, and 15%, respectively. The probability that you feel this sick if you have the flu, food poisoning. or the disease D, is 80%, 95%, 20%, respectively. What is the probability that you are sick because of food poisoning?
Let B be the event “you feel sick,” A 1 the event “you have the flu,” A 2 the event “you have some food poisoning,” and A 3 the event “you have some other disease.”
We are given P (A 1 ) = 0. 1 , P (A 2 ) = 0. 05 , P (A 3 ) = 0.15 and P [B|A 1 ] = 0. 8 , P [B|A 2 ] = 0. 95 , P [B|A 3 ] = 0 .2. We find the probability P [A 2 |B] by using Bayes’ rule:
P [A 2 |B] =
P (A 2 )P [B|A 2 ]
P (A 1 )P [B|A 1 ] + P (A 2 )P [B|A 2 ] + P (A 3 )P [B|A 3 ]
0. 05 × 0. 95
0. 1 × 0 .8 + 0. 05 × 0 .95 + 0. 15 × 0. 2
Can you find a probability space and events A, B so that
P [A|B] > P (A) and P [B|A] < P (B)?
The answer is no. Indeed, the first inequality implies
P (A ∩ B) > P (A)P (B)
while the second implies P (A ∩ B) < P (A)P (B),
and these are clearly incompatible.
Let X be a random variable that is uniform in [0, 1]. Calculate the variance of Xn^ for n ≥ 1.
We find
var(Xn) = E(X^2 n) − E(Xn)^2 =
2 n + 1
(n + 1)^2
We used the fact that
E(Xm) =
∫ (^1)
0
xmdx =
m + 1
State and prove Markov’s inequality.
Markov’s inequality states that if X is a nonnegative random variable, then
P (X ≥ a) ≤
E(X)
a
To prove the inequality, we observe that
X ≥ a 1 {X ≥ a}.
Taking the expectation, we get E(X) ≥ aP (X ≥ a).
Figure 3: The cpdf of X.
The random variable X has the c.p.d.f. shown above. Calculate E(X) and var(X).
We see that
E(X) =
∫ (^1)
− 1
x × 0. 125 dx + 1 × 0 .25 +
∫ (^2)
1
x × 0. 5 dx = 0 + 0.25 + 0.75 = 1.
Also,
E(X^2 ) =
∫ (^1)
− 1
x^2 × 0. 125 dx + 1 × 0 .25 +
∫ (^2)
1
x^2 × 0. 5 dx = 0. 25 /3 + 0.25 + (2^3 − 13 )0. 5 /3 = 1. 5.
Hence,
var(X) = 1. 5 − 12 = 0. 5.
