Suppose that the random variables Y1 and Y2 have joint probability distribution function.f(y1, y2) = 2, 0 ≤ y1 ≤ 1, 0 ≤ y2 ≤ 1, 0 ≤ y1 + y2 ≤ 1,0, elsewhere(a)Use R to calculate P(Y1 ≥ 1⁄6 | Y2 ≤ 1⁄5). (Round your answer to four decimal places.)P(Y1 ≥ 1⁄6 | Y2 ≤ 1⁄5) = (b)Use R to calculate P(Y1 ≥ 1⁄6 | Y2 = 1⁄5). (Round your answer to four decimal places.)P(Y1 ≥ 1⁄6 | Y2 = 1⁄5) = **Title: Understanding Joint Probability Distribution Functions****Description:**Suppose that the random variables \( Y_1 \) and \( Y_2 \) have a joint probability distribution function defined as:\[ f(y_1, y_2) = \begin{cases} 2, & 0 \leq y_1 \leq 1, \ 0 \leq y_2 \leq 1, \ 0 \leq y_1 + y_2 \leq 1, \\0, & \text{elsewhere}.\end{cases} \]**Tasks:**(a) **Use R to Calculate Conditional Probability**Calculate \( P(Y_1 \geq 1/6 \mid Y_2 \leq 1/5) \). Provide your answer rounded to four decimal places.\[ P(Y_1 \geq 1/6 \mid Y_2 \leq 1/5) = \] [Text box for answer input] [Incorrect symbol](b) **Use R to Calculate Conditional Probability (Alternate Scenario)**Calculate \( P(Y_1 \geq 1/6 \mid Y_2 = 1/5) \). Provide your answer rounded to four decimal places.\[ P(Y_1 \geq 1/6 \mid Y_2 = 1/5) = \] [Text box for answer input] [Incorrect symbol]**Explanation:**This exercise involves evaluating the conditional probabilities in a defined region of a joint probability distribution function. The function \( f(y_1, y_2) \) is defined over a triangular region where \( 0 \leq y_1 + y_2 \leq 1 \). Use R programming to compute these probabilities with precision for further understanding.**Note:**The scenarios presented in (a) and (b) involve different conditions for \( Y_2 \), showcasing how such alterations impact probability calculations.

The joint probability density function is given as,The marginal density function of y1 is,The…

Answered: Suppose that the random variables Y1…

Math

Probability

Suppose that the random variables Y1 and Y2 have joint probability distribution function. f(y1, y2) = 2, 0 ≤ y1 ≤ 1,

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.8: Probability

Problem 19E

See similar textbooks

Related questions

Q: An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" ()…

A: The question is about probability distribution Given : To find : Prob. P ( X = x ) of the prob.…

Q: Let the r.v. X assume the value 'r' with the probability law : P(X = r) = q-¹ p ; r = 1, 2, 3. Find…

A: The probability mass function of X is given as-

Q: A person leaves for work between 8:00 am and 8:30 am and takes between 40 and 50 minutes to get…

A: Given that -

Q: Let Y be the Exponential random variable withl > 0. Find the following probabilities. 1(1 3(1

A: Hello! As you have posted more than 3 sub parts, we are answering the first 3 sub-parts. In case…

Q: x is a uniform random variable in the interval (x1,x2). Find the expected value of x

Q: For the discrete random variable X, P(X = 1) = p = 1 – P(X = -1). Please find c so as to be E(cX) =…

Q: Let the random variable X denote the number of trials in ecess of r that are ed to achieve the r-th…

A: Given that, The random variable X as the number of trails in excess of r that are required to…

Q: fx, (x) = 0.1 0.95-r

A: The probability distribution of the random variable X1,X2 and X3 is given as- fX1x=5x×0.1x×0.95-x,…

Q: Suppose you select a random point inside the unit square, i.e. Ω={(x,y)|0≤x,y≤1}. Determine the…

Q: Consider a random variable X taking the values k1, k2, ·.. , kn E R with probability P1, P2, ; Pn E…

A: Formula is

Question

Suppose that the random variables

Y₁

and

Y₂

have joint probability distribution function.

f(y₁, y₂) =

2,	0 ≤ y₁ ≤ 1, 0 ≤ y₂ ≤ 1, 0 ≤ y₁ + y₂ ≤ 1,
0,	elsewhere

(a)

Use R to calculate

P(Y₁ ≥ 1⁄6 | Y₂ ≤ 1⁄5).

(Round your answer to four decimal places.)

P(Y₁ ≥ 1⁄6 | Y₂ ≤ 1⁄5) =

(b)

Use R to calculate

P(Y₁ ≥ 1⁄6 | Y₂ = 1⁄5).

(Round your answer to four decimal places.)

P(Y₁ ≥ 1⁄6 | Y₂ = 1⁄5) =

**Title: Understanding Joint Probability Distribution Functions**

**Description:**

Suppose that the random variables \( Y_1 \) and \( Y_2 \) have a joint probability distribution function defined as:

\[ f(y_1, y_2) = \begin{cases}
2, & 0 \leq y_1 \leq 1, \ 0 \leq y_2 \leq 1, \ 0 \leq y_1 + y_2 \leq 1, \\
0, & \text{elsewhere}.
\end{cases} \]

**Tasks:**

(a) **Use R to Calculate Conditional Probability**

Calculate \( P(Y_1 \geq 1/6 \mid Y_2 \leq 1/5) \). Provide your answer rounded to four decimal places.

\[ P(Y_1 \geq 1/6 \mid Y_2 \leq 1/5) = \] [Text box for answer input] [Incorrect symbol]

(b) **Use R to Calculate Conditional Probability (Alternate Scenario)**

Calculate \( P(Y_1 \geq 1/6 \mid Y_2 = 1/5) \). Provide your answer rounded to four decimal places.

\[ P(Y_1 \geq 1/6 \mid Y_2 = 1/5) = \] [Text box for answer input] [Incorrect symbol]

**Explanation:**

This exercise involves evaluating the conditional probabilities in a defined region of a joint probability distribution function. The function \( f(y_1, y_2) \) is defined over a triangular region where \( 0 \leq y_1 + y_2 \leq 1 \). Use R programming to compute these probabilities with precision for further understanding.

**Note:**
The scenarios presented in (a) and (b) involve different conditions for \( Y_2 \), showcasing how such alterations impact probability calculations.

Video Video

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1: Introduction

Step 2: Part(a)

Step 3: part(b)

Solution

Step by step

Solved in 4 steps with 27 images

SEE SOLUTION Check out a sample Q&A here

Similar questions

An ordinary die is rolled. Find the probability of each event. Rolling a number greater than 1 but less than 6
An ordinary die is rolled. Find the probability of each event. Rolling a number greater than 4
Variance of a uniform probability distribution is Var(X)=(a-b)^ 2 /12 where b>a True False
Explain Functions of Normally and Log-normally Distributed Random Variables
Pollution Carbon monoxide (CO) emissions for a certain kind of car vary with mean 2.9 g>mi and standard devia-tion 0.4 g>mi. A company has 80 of these cars in its fleet. Let y represent the mean CO level for the company’sfleet.a) What’s the approximate model for the distribution of y?Explain.b) Estimate the probability that y is between 3.0 and3.1 g>mi.c) There is only a 5% chance that the fleet’s mean COlevel is greater than what value?
QI\ A/ Consider a discrete random variable X whose pmf is given by: Find: 1- E[ X ] and Var( X )? f(x) = 2- standard deviation of X? B/ Let X be a binomial random variable with parameters n=6, p=0.4. Compute x=-1,0,1 otherwise Fx (4) ?
Find the mean and variance for the probability distribution given by f(x)= 2x k(k+1)* -,x=1,2,3,...k
a discrete probability distribution is defined by the function P(X = x) = (x^2 + x) / 20 ; x = 1, 2, 3 a) use this information to fill in the second column in the table. The additional columns are for you to use as needed when finishing other parts of this problem. There may be an extra column. x P ( X = x) 1 2 3 total b) describe how you can determine that the algebraic function does indeed define a discrete probability c) use another column on the table as needed to find the mean, also known as the expected value d) Use another column on the table as needed to find quantities needed to calculate the standard deviation.
The variance for the binomial probability distribution is var(x) = P(1 - P) var(x) = nP var(x) = n(1 - P) var(x) = nP(1 - P)
The discrete random variable X has a cumulative distribution function, F(z) = P(X < z), defined by F(x)=(¹^), z = 1 Note: Range of X = {1,2,3} Find the value of A. O a. 6 O b. 3/2 O c. 8 O d. 2/3 Oe. 1 Of 5 Og. None of these Oh. 9 4 = 1,2,3 www ndows
What is the probability that x is less than or equal to 2? What is the probability that is an even number? Determine the cumulative destribution function for the function F(3)
For discrete probability distribution x 0 1 2 3 4 5 6 7f 0 k 2k 2k 3k k^2 2k^2 7k^2+k The mean of the distribution is