Suppose X and Y are independent random variables and that X~ on (2,5). Calculate the probability that 2X + Y < 8. Exp(14) and Y is uniformly distributed Proposed solution: Because X and Y are independent, the joint p.d.f. is given by the product of the p.d.f. of X and of Y,
Suppose X and Y are independent random variables and that X~ on (2,5). Calculate the probability that 2X + Y < 8. Exp(14) and Y is uniformly distributed Proposed solution: Because X and Y are independent, the joint p.d.f. is given by the product of the p.d.f. of X and of Y,
A First Course in Probability (10th Edition)
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![**Problem Statement:**
Suppose \( X \) and \( Y \) are independent random variables and that \( X \sim \text{Exp}(14) \) and \( Y \) is uniformly distributed on (2, 5). Calculate the probability that \( 2X + Y < 8 \).
**Proposed Solution:**
Because \( X \) and \( Y \) are independent, the joint probability density function (p.d.f.) is given by the product of the p.d.f. of \( X \) and \( Y \), that is
\[
f_{X,Y}(x, y) =
\begin{cases}
\frac{1}{3} e^{-14x} & \text{if } x > 0, y > 0 \\
0 & \text{otherwise.}
\end{cases}
\]
**Solution Details:**
The desired probability is calculated as follows:
\[
\mathbb{P}(2X + Y < 8) = \int_{4}^{10} \int_{0}^{8-y} \frac{1}{3} e^{-14s} \, ds \, dt = \int_{4}^{10} \left[ -\frac{1}{42} e^{-14s} \right]_{0}^{8-t} \, dt
\]
This simplifies to:
\[
= \int_{4}^{10} \frac{14}{3} \left( 1 - e^{-14(8-t)} \right) dt = \frac{14}{3} \left[ t - \frac{1}{14} e^{-14(8-t)} \right]_{4}^{10}
\]
Evaluating this expression results in:
\[
= 28 - \frac{1}{14} (e^{-56} - e^{-28}).
\]
This provides the solution to the probability problem using integration of the joint probability density function over the specified limits.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9a42504d-f572-4a8c-acec-8d504fa953af%2F1b4f6d57-2fb0-467e-b6cf-6eb55175288a%2Fvz3osy_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Suppose \( X \) and \( Y \) are independent random variables and that \( X \sim \text{Exp}(14) \) and \( Y \) is uniformly distributed on (2, 5). Calculate the probability that \( 2X + Y < 8 \).
**Proposed Solution:**
Because \( X \) and \( Y \) are independent, the joint probability density function (p.d.f.) is given by the product of the p.d.f. of \( X \) and \( Y \), that is
\[
f_{X,Y}(x, y) =
\begin{cases}
\frac{1}{3} e^{-14x} & \text{if } x > 0, y > 0 \\
0 & \text{otherwise.}
\end{cases}
\]
**Solution Details:**
The desired probability is calculated as follows:
\[
\mathbb{P}(2X + Y < 8) = \int_{4}^{10} \int_{0}^{8-y} \frac{1}{3} e^{-14s} \, ds \, dt = \int_{4}^{10} \left[ -\frac{1}{42} e^{-14s} \right]_{0}^{8-t} \, dt
\]
This simplifies to:
\[
= \int_{4}^{10} \frac{14}{3} \left( 1 - e^{-14(8-t)} \right) dt = \frac{14}{3} \left[ t - \frac{1}{14} e^{-14(8-t)} \right]_{4}^{10}
\]
Evaluating this expression results in:
\[
= 28 - \frac{1}{14} (e^{-56} - e^{-28}).
\]
This provides the solution to the probability problem using integration of the joint probability density function over the specified limits.
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