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Probability Theory : Conditional Distributions and Independence

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Example Questions

Example Question #1 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - 2 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

Possible Answers:

c = 1

c = 6

$c = \frac{3}{2}$

c = 12

c = 2

Correct answer:

c = 6

Explanation:

In order to find the value of $\uptext{c}$ , we need to take find the double integral of the function

Let's find what the bounds are for both $\uptext{x}$ , and $\uptext{y}$

We look at the p.d.f to see that the bounds for $\uptext{x}$ are, 0<x<y , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 2 y}\, dx\, dy$

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 2 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} c e^{- 2 y} - c e^{- 3 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{2} e^{- 2 y} + \frac{c}{3} e^{- 3 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{6}\right) + \lim_{b \to \infty}\left(- \frac{c}{2} e^{- 2 b} + \frac{c}{3} e^{- 3 b}\right)$

$\frac{c}{6} = 1$

Now we simply solve for $\uptext{c}$

c = 6

Report an Error

Example Question #2 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - 19 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

Possible Answers:

c = 380

$c = \frac{1}{380}$

c = 400

$c = \frac{1}{400}$

c = 360

Correct answer:

c = 380

Explanation:

In order to find the value of $\uptext{c}$ , we need to take find the double integral of the function.

Let's find what the bounds are for both $\uptext{x}$ , and $\uptext{y}$ .

We look at the p.d.f to see that the bounds for $\uptext{x}$ are, 0<x<y , and for $\uptext{y}$ , $0<y<\infty$ .

Now let's set up the double integral.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 19 y}\, dx\, dy$

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - 19 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} c e^{- 19 y} - c e^{- 20 y}\, dy = 1$
To evaluate this, we need to use the limit definition

$\lim_{b\to \infty}\left(- \frac{c}{19} e^{- 19 y} + \frac{c}{20} e^{- 20 y}\right) \Big|_{ 0 }^{ 19 }$

$- \lim_{b \to \infty}\left(- \frac{c}{380}\right) + \lim_{b\to \infty}\left(- \frac{c}{19 e^{361}} + \frac{c}{20 e^{380}}\right)$

$\frac{c}{380} = 1$

Now we simply solve for $\uptext{c}$

c = 380

Report an Error

Example Question #3 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 7 x - 14 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

Possible Answers:

c = 294

c = 588

c = 1

$c = \frac{147}{2}$

c = 98

Correct answer:

c = 294

Explanation:

In order to find the value of $\uptext{c}$ , we need to take find the double integral of the function.

Let's find what the bounds are for both $\uptext{x}$ , and $\uptext{y}$ .

We look at the p.d.f to see that the bounds for $\uptext{x}$ are, 0<x<y , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 7 x - 14 y}\, dx\, dy$
Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 7 x - 14 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{7} e^{- 14 y} - \frac{c}{7} e^{- 21 y}\, dy = 1$

To evaluate this, we need to use the limit definition
$\lim_{b \to \infty}\left(- \frac{c}{98} e^{- 14 y} + \frac{c}{147} e^{- 21 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{294}\right) + \lim_{b \to \infty}\left(- \frac{c}{98} e^{- 14 b} + \frac{c}{147} e^{- 21 b}\right)$

$\frac{c}{294} = 1$

Now we simply solve for $\uptext{c}$

c = 294

Report an Error

Example Question #4 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 5 x - 6 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

Possible Answers:

c = 66

c = 1

c = 132

$c = \frac{33}{2}$

c = 22

Correct answer:

c = 66

Explanation:

In order to find the value of $\uptext{c}$ , we need to take find the double integral of the function

Let's find what the bounds are for both $\uptext{x}$ , and $\uptext{y}$

We look at the p.d.f to see that the bounds for $\uptext{x}$ are, 0<x<y , and for $\uptext{y}$ , $0<y<\infty$

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 5 x - 6 y}\, dx\, dy$
Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 5 x - 6 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{5} e^{- 6 y} - \frac{c}{5} e^{- 11 y}\, dy = 1$
To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{30} e^{- 6 y} + \frac{c}{55} e^{- 11 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{66}\right) + \lim_{b \to \infty}\left(- \frac{c}{30} e^{- 6 b} + \frac{c}{55} e^{- 11 b}\right)$

$\frac{c}{66} = 1$

Now we simply solve for $\uptext{c}$

c = 66

Report an Error

Example Question #5 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 9 x - 12 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

Possible Answers:

c = 504

c = 252

c = 84

c = 1

c = 63

Correct answer:

c = 252

Explanation:

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 9 x - 12 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{9} e^{- 12 y} - \frac{c}{9} e^{- 21 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{108} e^{- 12 y} + \frac{c}{189} e^{- 21 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{252}\right) + \lim_{b \to \infty}\left(- \frac{c}{108} e^{- 12 b} + \frac{c}{189} e^{- 21 b}\right)$

$\frac{c}{252} = 1$

Now we simply solve for $\uptext{c}$

c = 252

Report an Error

Example Question #1 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 6 x - 10 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

Possible Answers:

$c = \frac{160}{3}$

c = 1

c = 160

c = 320

c = 40

Correct answer:

c = 160

Explanation:

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 6 x - 10 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{6} e^{- 10 y} - \frac{c}{6} e^{- 16 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{60} e^{- 10 y} + \frac{c}{96} e^{- 16 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{160}\right) + \lim_{b \to \infty}\left(- \frac{c}{60} e^{- 10 b} + \frac{c}{96} e^{- 16 b}\right)$

$\frac{c}{160} = 1$

Now we simply solve for $\uptext{c}$

c = 160

Report an Error

Example Question #7 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 10 x - 13 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$

Possible Answers:

$c = \frac{299}{4}$

c = 299

$c = \frac{299}{3}$

c = 598

c = 1

Correct answer:

c = 299

Explanation:

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 10 x - 13 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{10} e^{- 13 y} - \frac{c}{10} e^{- 23 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{130} e^{- 13 y} + \frac{c}{230} e^{- 23 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{299}\right) + \lim_{b \to \infty}\left(- \frac{c}{130} e^{- 13 b} + \frac{c}{230} e^{- 23 b}\right)$

$\frac{c}{299} = 1$

Now we simply solve for $\uptext{c}$

c = 299

Report an Error

Example Question #1 : Conditional Distributions And Independence

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 8 x - 9 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

Possible Answers:

c = 1

c = 51

c = 306

$c = \frac{153}{4}$

c = 153

Correct answer:

c = 153

Explanation:

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 8 x - 9 y}\, dx\, dy$
Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 8 x - 9 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{8} e^{- 9 y} - \frac{c}{8} e^{- 17 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{72} e^{- 9 y} + \frac{c}{136} e^{- 17 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{153}\right) + \lim_{b \to \infty}\left(- \frac{c}{72} e^{- 9 b} + \frac{c}{136} e^{- 17 b}\right)$

$\frac{c}{153} = 1$

Now we simply solve for $\uptext{c}$

c = 153

Report an Error

Example Question #1 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- x - y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

Possible Answers:

$c = \frac{1}{2}$

c = 2

$c = \frac{2}{3}$

c = 1

c = 4

Correct answer:

c = 2

Explanation:

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- x - y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} c e^{- y} - c e^{- 2 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- c e^{- y} + \frac{c}{2} e^{- 2 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{2}\right) + \lim_{b \to \infty}\left(- c e^{- b} + \frac{c}{2} e^{- 2 b}\right)$

$\frac{c}{2} = 1$

Now we simply solve for $\uptext{c}$

c = 2

Report an Error

Example Question #10 : Multiple Random Variables

Let $\uptext{X}$ , and $\uptext{Y}$ be the lifespans (in hours) of two electronic devices, and their joint probability mass function is given below.

$\operatorname{f_{X,Y}}{\left (x,y \right )} = \begin{cases} c e^{- 2 x - 10 y} & 0<x<y<\infty \\ 0, & \text{Otherwise} \end{cases}$

Determine the value of $\uptext{c}$ .

Possible Answers:

c = 30

c = 240

c = 120

c = 40

c = 1

Correct answer:

c = 120

Explanation:

Now let's set up the double integral

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 2 x - 10 y}\, dx\, dy$

Before we evaluate it, we need to remember to set the double integral equal to one, since we are essentially solving for the c.d.f.

$\int_{0}^{\infty}\int_{0}^{y} c e^{- 2 x - 10 y}\, dx\, dy = 1$

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{2} e^{- 10 y} - \frac{c}{2} e^{- 12 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{20} e^{- 10 y} + \frac{c}{24} e^{- 12 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{120}\right) + \lim_{b \to \infty}\left(- \frac{c}{20} e^{- 10 b} + \frac{c}{24} e^{- 12 b}\right)$

$\frac{c}{120} = 1$

Now we simply solve for $\uptext{c}$

c = 120

Report an Error

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Probability Theory : Conditional Distributions and Independence

Study concepts, example questions & explanations for Probability Theory

All Probability Theory Resources

Example Questions

Example Question #1 : Multiple Random Variables

Example Question #2 : Multiple Random Variables

Example Question #3 : Multiple Random Variables

Example Question #4 : Multiple Random Variables

Example Question #5 : Multiple Random Variables

Example Question #1 : Multiple Random Variables

Example Question #7 : Multiple Random Variables

Example Question #1 : Conditional Distributions And Independence

Example Question #1 : Multiple Random Variables

Example Question #10 : Multiple Random Variables

All Probability Theory Resources

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