Probability Theory : Probability Theory

Study concepts, example questions & explanations for Probability Theory

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Example Questions

Example Question #1 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}

Possible Answers:

\displaystyle c = 1

\displaystyle c = 6

\displaystyle c = 12

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

\displaystyle c = 2

Correct answer:

\displaystyle c = 6

Explanation:

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

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

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

Now let's set up the double integral

\displaystyle \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.

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

Now evaluate the double integral

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

To evaluate this, we need to use the limit definition

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

\displaystyle - \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)

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

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

\displaystyle c = 6

Example Question #2 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}

Possible Answers:

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

\displaystyle c = 380

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

\displaystyle c = 400

\displaystyle c = 360

Correct answer:

\displaystyle c = 380

Explanation:

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

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

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

Now let's set up the double integral.

\displaystyle \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.

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

 

Now evaluate the double integral

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

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

\displaystyle - \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)

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

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

\displaystyle c = 380

 

Example Question #3 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}.

Possible Answers:

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

\displaystyle c = 294

\displaystyle c = 588

\displaystyle c = 98

\displaystyle c = 1

Correct answer:

\displaystyle c = 294

Explanation:

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

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

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

Now let's set up the double integral

\displaystyle \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.

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


Now evaluate the double integral

\displaystyle \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
\displaystyle \lim_{b \to \infty}\left(- \frac{c}{98} e^{- 14 y} + \frac{c}{147} e^{- 21 y}\right) \Big|_{ 0 }^{ b }

\displaystyle - \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)

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


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

\displaystyle c = 294

Example Question #4 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}.

Possible Answers:

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

\displaystyle c = 132

\displaystyle c = 66

\displaystyle c = 1

\displaystyle c = 22

Correct answer:

\displaystyle c = 66

Explanation:

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

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

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

Now let's set up the double integral

\displaystyle \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.

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


Now evaluate the double integral

\displaystyle \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

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

\displaystyle - \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)

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

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

\displaystyle c = 66

Example Question #5 : Multiple Random Variables

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


\displaystyle \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 \displaystyle \uptext{c}

Possible Answers:

\displaystyle c = 504

\displaystyle c = 84

\displaystyle c = 1

\displaystyle c = 63

\displaystyle c = 252

Correct answer:

\displaystyle c = 252

Explanation:

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

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

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

Now let's set up the double integral

\displaystyle \int_{0}^{\infty}\int_{0}^{y} c e^{- 9 x - 12 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.

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


Now evaluate the double integral

\displaystyle \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

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

\displaystyle - \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)

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

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

\displaystyle c = 252

Example Question #6 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}.

Possible Answers:

\displaystyle c = 1

\displaystyle c = 320

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

\displaystyle c = 40

\displaystyle c = 160

Correct answer:

\displaystyle c = 160

Explanation:

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

Let's find what the bounds are for both \displaystyle \uptext{x}, and \displaystyle \uptext{y}
We look at the p.d.f to see that the bounds for \displaystyle \uptext{x} are, \displaystyle 0< x< y, and for \displaystyle \uptext{y}, \displaystyle 0< y< \infty

Now let's set up the double integral

\displaystyle \int_{0}^{\infty}\int_{0}^{y} c e^{- 6 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.


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


Now evaluate the double integral

\displaystyle \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

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

\displaystyle - \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)


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


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

\displaystyle c = 160

Example Question #7 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}

Possible Answers:

\displaystyle c = 1

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

\displaystyle c = 598

\displaystyle c = 299

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

Correct answer:

\displaystyle c = 299

Explanation:

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

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

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

Now let's set up the double integral

\displaystyle \int_{0}^{\infty}\int_{0}^{y} c e^{- 10 x - 13 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.

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


Now evaluate the double integral


\displaystyle \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

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

\displaystyle - \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)


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

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

\displaystyle c = 299

Example Question #8 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}.

Possible Answers:

\displaystyle c = 1

\displaystyle c = 153

\displaystyle c = 306

\displaystyle c = 51

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

Correct answer:

\displaystyle c = 153

Explanation:

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

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

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

Now let's set up the double integral


\displaystyle \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.

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

Now evaluate the double integral

\displaystyle \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

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

\displaystyle - \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)

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


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


\displaystyle c = 153

Example Question #9 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}.

Possible Answers:

\displaystyle c = 4

\displaystyle c = 2

\displaystyle c = 1

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

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

Correct answer:

\displaystyle c = 2

Explanation:

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

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

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

Now let's set up the double integral

\displaystyle \int_{0}^{\infty}\int_{0}^{y} c e^{- x - 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.

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

Now evaluate the double integral

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


To evaluate this, we need to use the limit definition

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

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

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

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

\displaystyle c = 2

Example Question #10 : Multiple Random Variables

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

\displaystyle \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 \displaystyle \uptext{c}.

Possible Answers:

\displaystyle c = 120

\displaystyle c = 30

\displaystyle c = 40

\displaystyle c = 240

\displaystyle c = 1

Correct answer:

\displaystyle c = 120

Explanation:

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

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

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


Now let's set up the double integral

\displaystyle \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.

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


Now evaluate the double integral


\displaystyle \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

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

\displaystyle - \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)

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

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

\displaystyle c = 120

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