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All Probability Theory Resources

3 Practice Tests Question of the Day Flashcards Learn by Concept

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Flashcards: Probability Theory

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

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

Determine the value of $\uptext{c}$

c = 400

c = 200

$c = \frac{200}{3}$

c = 1

c = 50

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Flashcards: Probability Theory

Determine the value of $\uptext{c}$

Your answer:

c = 400

Your answer:

$c = \frac{200}{3}$

Your answer:

c = 1

Your answer:

c = 50

Correct answer:

c = 200

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

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

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.

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

Now evaluate the double integral

$\int_{0}^{\infty} \frac{c}{10} e^{- 10 y} - \frac{c}{10} e^{- 20 y}\, dy = 1$

To evaluate this, we need to use the limit definition

$\lim_{b \to \infty}\left(- \frac{c}{100} e^{- 10 y} + \frac{c}{200} e^{- 20 y}\right) \Big|_{ 0 }^{ b }$

$- \lim_{b \to \infty}\left(- \frac{c}{200}\right) + \lim_{b \to \infty}\left(- \frac{c}{100} e^{- 10 b} + \frac{c}{200} e^{- 20 b}\right)$

$\frac{c}{200} = 1$

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

c = 200

Flashcards: Probability Theory

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 $\uptext{c}$

c = 1

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

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

c = 299

c = 598

Report an Error Copyright Notice

Flashcards: Probability Theory

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 $\uptext{c}$

Your answer:

c = 1

Your answer:

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

Your answer:

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

Your answer:

c = 598

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

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All Probability Theory Resources

Flashcards: Probability Theory

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All Probability Theory Resources

Flashcards: Probability Theory

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Flashcards: Probability Theory

Flashcards: Probability Theory

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