From a sack of fruit containing 3 oranges, 2 apples, and 3 bananas, a random sample of 4 pieces of fruit is selected. Suppose X is the number of oranges and Y is the number of apples in the sample. (a) Find the joint probability distribution of X and Y. (b) Find P[(X,Y)EA], where A is the region that is given by {(x,y) | x+y<3}. C (a) Complete the joint probability distribution below. (Type integers or simplified fractions.) X f(x,y) 0 3 4 y 0 2

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### Joint Probability Distribution of Oranges and Apples in a Random Sample

#### Problem Statement
From a sack of fruit containing 3 oranges, 2 apples, and 3 bananas, a random sample of 4 pieces of fruit is selected. Suppose \( X \) is the number of oranges and \( Y \) is the number of apples in the sample.

(a) Find the joint probability distribution of \( X \) and \( Y \).  
(b) Find \( P((X,Y) \in A) \), where \( A \) is the region that is given by \(\{(x,y) | x + y \leq 3\}\).

#### Solution

(a) **Complete the joint probability distribution below.**

| \( f(x,y) \) | \( y \\ x \) | 0 | 1 | 2 | 3 | 4 |
|--------------|--------------|---|---|---|---|---|
| 0            |              |   |   |   |   |   |

(b) **Finding \( P((X,Y) \in A) \):**

To solve for the probability, one must integrate or sum over the joint distribution \( f(x,y) \) for all \((x,y)\) in the region \( A \).
Transcribed Image Text:### Joint Probability Distribution of Oranges and Apples in a Random Sample #### Problem Statement From a sack of fruit containing 3 oranges, 2 apples, and 3 bananas, a random sample of 4 pieces of fruit is selected. Suppose \( X \) is the number of oranges and \( Y \) is the number of apples in the sample. (a) Find the joint probability distribution of \( X \) and \( Y \). (b) Find \( P((X,Y) \in A) \), where \( A \) is the region that is given by \(\{(x,y) | x + y \leq 3\}\). #### Solution (a) **Complete the joint probability distribution below.** | \( f(x,y) \) | \( y \\ x \) | 0 | 1 | 2 | 3 | 4 | |--------------|--------------|---|---|---|---|---| | 0 | | | | | | | (b) **Finding \( P((X,Y) \in A) \):** To solve for the probability, one must integrate or sum over the joint distribution \( f(x,y) \) for all \((x,y)\) in the region \( A \).
### Joint Density Function and Marginal Density

#### Introduction
A fast-food restaurant operates both a drive-through facility and a walk-in facility. On a randomly selected day, let \(X\) and \(Y\) represent the proportions of the time that the drive-through and walk-in facilities are in use, respectively. The joint density function of these random variables is given by:

\[ 
f(x, y) = 
\begin{cases} 
\frac{2}{11} (6x + 5y), & 0 \leq x \leq 1, \ 0 \leq y \leq 1 \\
0, & \text{elsewhere}
\end{cases} 
\]

#### Problem Statement
Complete parts (a) through (c).

#### Part (a): Marginal Density of \(X\)

##### Task
Find the marginal density of \(X\).

##### Instructions
Select the correct choice below and fill in the answer box to complete your choice.

##### Choices
- \( A. \ h(y) = \Box, \ \text{for} \ 0 \leq y \leq 1 \)
- \( B. \ g(x) = \Box, \ \text{for} \ 0 \leq x \leq 1 \)

---

### Detailed Explanation
- To find the marginal density function \( g(x) \) of \(X\), you integrate the joint density \( f(x, y) \) over all possible values of \( y \).

\[ g(x) = \int_{0}^{1} f(x, y) \, dy \]

- Similarly, to find the marginal density function \( h(y) \) of \(Y\), you integrate the joint density \( f(x, y) \) over all possible values of \( x \).

\[ h(y) = \int_{0}^{1} f(x, y) \, dx \]

This problem challenges students to use their knowledge of joint density functions and marginal distributions to derive the marginal density of a given variable.
Transcribed Image Text:### Joint Density Function and Marginal Density #### Introduction A fast-food restaurant operates both a drive-through facility and a walk-in facility. On a randomly selected day, let \(X\) and \(Y\) represent the proportions of the time that the drive-through and walk-in facilities are in use, respectively. The joint density function of these random variables is given by: \[ f(x, y) = \begin{cases} \frac{2}{11} (6x + 5y), & 0 \leq x \leq 1, \ 0 \leq y \leq 1 \\ 0, & \text{elsewhere} \end{cases} \] #### Problem Statement Complete parts (a) through (c). #### Part (a): Marginal Density of \(X\) ##### Task Find the marginal density of \(X\). ##### Instructions Select the correct choice below and fill in the answer box to complete your choice. ##### Choices - \( A. \ h(y) = \Box, \ \text{for} \ 0 \leq y \leq 1 \) - \( B. \ g(x) = \Box, \ \text{for} \ 0 \leq x \leq 1 \) --- ### Detailed Explanation - To find the marginal density function \( g(x) \) of \(X\), you integrate the joint density \( f(x, y) \) over all possible values of \( y \). \[ g(x) = \int_{0}^{1} f(x, y) \, dy \] - Similarly, to find the marginal density function \( h(y) \) of \(Y\), you integrate the joint density \( f(x, y) \) over all possible values of \( x \). \[ h(y) = \int_{0}^{1} f(x, y) \, dx \] This problem challenges students to use their knowledge of joint density functions and marginal distributions to derive the marginal density of a given variable.
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