Question 1. Question 4.

İ s t a n b u l K ü l t ü r U n i v e r s i t y Faculty of Engineering

MCB1007

Introduction to Probability and Statistics Second Midterm

Fall 2013-2014

Number:

Name:

Department:

Section:

 – You have 90 minutes to complete the exam. Please do not leave the examination room in the first 30 minutes of the exam. There are six questions, of varying credit (100 points total). Indicate clearly your final answer to each question. You are allowed to use a calculator. During the exam, please turn off your cell phone(s).

You cannot use the book or your notes. You have one page for “cheat-sheet” notes.

The answer key to this exam will be posted on Department of Mathematics and Computer Science board after the exam.

Good luck! Emel Yavuz Duman, PhD.

M. Fatih Uçar, PhD.

Question 1. Question 4.

Question 2. Question 5.

Question 3. Question 6.

TOTAL

  10 + 10 points Suppose that a bag has 3 boxes, one of them containing 2 bullets and two of them containing 1 bullet. A person is asked to randomly choose a box and than shoot a target until the last bullet is fired. Suppose 90% for this person, the bullet will hit the target. Let X be the number of bullets hitting the target and Y be the number of bullets in the selected box that the person has.

(a) Find the joint probability distribution of X and Y .

Answer. For the values of X = 0, 1, 2 and Y = 1, 2 we have the following probabilities:

f(0, 1) = 2 3

1 0



0.9⁰0.1¹ = 1

15 = 20

300, f(0, 2) = 1 3

2 0



0.9⁰0.1² = 1 300

f(1, 1) = 2 3

1 1



0.9¹0.1⁰ = 3

5 = 180

300, f(1, 2) = 1 3

2 1



0.9¹0.1¹ = 3

50 = 18 300 f(2, 1) = 0, f(2, 2) = 1

3

2 2



0.9²0.1⁰ = 27

100 = 81 300 Therefore, the joint probability distribution of X and Y is

HHy HHHH

x 0 1 2 h(y)

1 20/300 180/300 0 200/300

2 1/300 18/300 81/300 100/300 g(x) 21/300 198/300 81/300 1

(b) Find the conditional variance of X given Y = 1.

Answer. Conditional distribution of X given Y = 1 is

f(x|1) = f(x, 1) h(1) = 3

2f(x, 1) =

⎧⎪

⎨

⎪⎩

32f(0, 1) = ³₂₃₀₀²⁰ = ₁₀¹ for x= 0

32f(1, 1) = ³₂¹⁸⁰₃₀₀ = ₁₀⁹ for x= 1

32f(2, 1) = ³₂0 = 0 for x= 2 On the other hand, since

E[X|1] =

2 x=0

xf(x|1) = 0f(0|1) + 1f(1|1) + 2f(2|1) = 9 10 and

E[X²|1] =²

x=0

x²f(x|1) = 0²f(0|1) + 1²f(1|1) + 2²f(2|1) = 9 10 Thus the conditional variance of X given Y = 1 is

σ²_X|1 = E[(X − μX|1)²|1] = E[X²|1] − (E[X|1])² = 9 10 −

 9 10

₂

= 9 100.

 8 + 7 points (a) The moment-generating function of the random variable X is M_X(t). If we let

R_X(t) = ln M_X(t), show that _dt^dR_X(t)

t=0= μ and _dt^d2₂R_X(t)

t=0= σ².

Answer.

Hint: ^d

dxln u(x) = ^u_u(x)^(x) It is easy to see that

d

dtR_X(t)

t=0 = d

dtln M_X(t)

t=0 = M_X^ (t) M_X(t)

t=0= M_X^ (0) M_X(0) = μ^₁

1 = μ, and

d²

dt²R_X(t)

t=0= d dt

M_X^ (t) M_X(t)

t=0 = M_X^(t)M_X(t) − (M_X^ (t))² (M_X(t))²

t=0

= M_X^(0)M_X(0) − (M_X^ (0))²

(M_X(0))² = μ^₂− μ² 1 = σ².

(b) Use the results given in (a) to find the mean and the variance of X having the moment- generating function M_X(t) = e^4(et−1).

Answer. Since

M_X(t) = e^4(et−1) ⇒ RX(t) = log M_X(t) = 4(e^t− 1) then, we see that

μ= d

dt(4(e^t− 1))

t=0

= 4e^t

t=0

= 4, and

σ² = d²

dt²(4(e^t− 1)) = 4e^t

t=0 = 4.

 8 + 7 points A random variable X has the density function given by

f(x) =

6x(1 − x) for 0 < x < 1

0 elsewhere.

(a) Find P(|X − μ| ≥ 0.7).

Answer. If we use the definition of a mean, we obtain that

μ= E[X] =

 _∞

−∞

xf(x)dx =

 ₁

0

6x²(1 − x)dx = 6

 ₁

0

(x²− x³)dx

= 6

x³ 3 −

x⁴ 4

₁

0

= 6

1 3 −

1 4



= 1

2 = 0.5.

On the other hand, since

P(|X − μ| ≥ 0.7) = P (|X − 0.5| ≥ 0.7) = 1 − P (|X − 0.5| < 0.7), and

|X−0.5| < 0.7 ⇔ −0.7 < X−0.5 < 0.7 ⇔ −0.7+0.5 < X < 0.7+0.5 ⇔ −0.2 < X < 1.2, thus

P(|X−0.5| ≥ 0.7) = 1−P (|X−0.5| < 0.7) = 1−

 _1.2

−0.2

f(x)dx = 1−

 ₁

0

f(x)dx = 1−1 = 0.

(b) Find an upper bound for P(|X − 0.5| ≥ 0.7) and compare it with the result in (a).

Answer. We need to use the Chebyshev’s inequality given by P(|X − μ| ≥ ε) ≤ σ²/ε²

where mean μ = E(X) = 0.5 and ε = 0.7. Variance of X can be calculated by using the formula σ² = E(X²) − [E(X)]². Since

E[X²] =

 _∞

−∞

x²f(x)dx =

 ₁

0

6x³(1 − x)dx = 6

 ₁

0

(x³ − x⁴)dx

= 6

x⁴ 4 −

x⁵ 4

₁

0

= 6

1 4 −

1 5



= 3

10 = 0.3 then

σ² = E(X²) − [E(X)]² = 0.3 − 0.5² = 0.05.

So, an upper bound for the given probability is P(|X − 0.5| ≥ 0.7) ≤ 0.05

0.7² = 0.1020.

Therefore, we can say that an upper bound for the given probability is 0.1020 which is

 10 + 10 points A doctor wishes to perform an experiment on5 patients. Suppose that the probability that a randomly selected patient agrees to participate to this experiment is 0.2.

(a) What is the probability that 15 patients must be asked before 5 are found who agree to participate?

Answer. X, the number of couples who agree to participate, is a negative-binomial random variable. So,

b^∗(15; 5, 0.2) =

14 4



(0.2)⁵(0.8)¹⁰= 0.03439

(b) What is the probability that at least 4 patients must be asked before the first patient is found who agree to participate?

Answer. X, the number of couples who agree to participate, is a geometric random variable. So,

P(X ≥ 4) = 1 − P (X < 4) = 1 − [g(1; 0.2) + g(2; 0.2) + g(3; 0.2)]

= 1 −

(0.2)(0.8)⁰+ (0.2)(0.8)¹+ (0.2)(0.8)²

= 1 − 0.488 = 0.512

 15 points

Suppose you roll a pair of non-fair, six sided dice 144 times, which are loaded in such a way that each odd number is three times as likely to occur as each even number. Let the random variable X denote the sum of the points on both dice which can come up in one roll. Use a Poisson approximation to determine the probability of observing that X = 6 exactly four times in this 144 rolls.

Answer. Since S = {1, 2, 3, 4, 5, 6}, and each odd number is three times as likely to occur as each even number, we have

P(X = 2) = P (X = 4) = P (X = 6) = p and P (X = 1) = P (X = 3) = P (X = 5) = 3p.

So, we get

P(X = 1)+P (X = 2)+P (X = 3)+P (X = 4)+P (X = 5)+P (X = 6) = 12p = 1 ⇒ p = 1 12. Let we define A= {sum of the points on both dice which can come up is 6}. Thus, we have A= {(4, 2), (2, 4), (5, 1), (1, 5), (3, 3)}. Therefore, we get

P(A) = 1 12

1 12+ 1

12 1 12+ 3

12 3 12 + 3

12 3 12 + 3

12 3

12 = 29 144 = θ.

Hence, using Poisson approximation we obtain that the probability of observing that X = 6 exactly four times in the 144 rolls with the parameter λ = nθ = 144 ·₁₄₄²⁹ = 29 is

p(4; 29) = 29⁴e⁻²⁹

4! = 7.496 × 10⁻⁹.

 15 points Suppose that there are two types of airplanes, one of them has two engines and other one has four engines. It is given that an airplane engine will fail, during the flight, with probability1 −θ (where θ = 1), independently from engine to engine; and suppose that the airplane will make a successful flight if at least 50 percent of its engines remain operative. For what values of θ is a two-engine plane preferable to a four-engine plane?

Answer. As each engine is assumed to fail or function independently of what happens with the other engines, it follows that X, the number of engines remaining operative, is a binomial random variable. Hence, the probability that a four-engine plane makes a successful flight is

P(X ≥ 2) = 1 − P (X < 2) = 1 − (P (X = 0) + P (X = 1))

= 1 −

4 0



θ⁰(1 − θ)⁴⁻⁰+

4 1



θ¹(1 − θ)⁴⁻¹



= 1 − (1 − θ)⁴− 4θ(1 − θ)³ =

whereas the corresponding probability for a two-engine plane is P(X ≥ 1) = 1 − P (X < 1) = 1 − P (X = 0) = 1 −

2 0



θ⁰(1 − θ)²⁻⁰ = 1 − (1 − θ)² Hence, the two-engine plane is safer if

1 − (1 − θ)² ≥ 1 − (1 − θ)⁴− 4θ(1 − θ)³ ⇔ (1 − θ)² ≤ (1 − θ)⁴+ 4θ(1 − θ)³ ⇔ 1 ≤ (1 − θ)²+ 4θ(1 − θ) = 1 − 2θ + θ²+ 4θ − 4θ² ⇔ 0 ≤ 2θ − 3θ² ⇔

0 ≤ 2θ − 3θ² = θ(2 − 3θ) ⇔ 0 ≤ 2 − 3θ ⇔ θ ≤ 2 3