5 pair of RVs

1 / 10

# 5 pair of RVs - PowerPoint PPT Presentation

5 pair of RVs. 5-1: joint pmf. In a box are three dice. Die 1 is normal; die 2 has no 6 face, but instead two 5 faces; die 3 has no 5 face, but instead two 6 faces. The experiment consists of selecting a die at random, followed by a toss with that die.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about '5 pair of RVs' - conor

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### 5 pair of RVs

5-1: joint pmf
• In a box are three dice. Die 1 is normal; die 2 has no 6 face, but instead two 5 faces; die 3 has no 5 face, but instead two 6 faces.
• The experiment consists of selecting a die at random, followed by a toss with that die.
• Let X be the die number that is selected, and let Y be the face value of that die.
• Find P(X = x, Y = y) for all possible x and y.
5-2: Joint pmf
• A packet switch has two input ports and two output ports. At a given time slot, a packet arrives at each input port with probability and is equally likely to be destined to output port 1 or 2. Let X and Y be the number of packets destined for output ports 1 and 2, respectively. Find the pmf of X and Y.
• Three outcomes for an input port can take the following values: (i) “n”, no packet arrival; (ii) “a1”, packet arrival destined for output port 1; (iii) “a2”, packet arrival destined for output port 2.
5-3: marginal pmf
• We pick a message, whose length N follows a geometric distribution with parameter 1-p and SN={0,1,2,…}.
• Find the joint pmf and the marginal pmf’s of Q and R, where Q is the quotient in the division of N by constant M, and R is the number of remaining bytes.
5-5: joint pdf
• Joint pdf is given by:
• Find c.
• Find marginal pdf’s
• Find P[X+Y1]
5-7: condi. prob. (discrete)
• The total number of defects X on a chip is a Poisson random variable with mean .
• Each defect has a probability p of falling in a specific region R and the location of each defect is independent of the locations of other defects.
• Find the pmf of the number of defects Y that fall in the region R.
5-8: condi. Prob. (conti.)
• X is selected at random from the unit interval; Y is then selected at random from the interval (0, X).
• Find the cdf of Y.
5-9: function of RVs
• A system with standby redundancy has a single key component in operation and a duplicate of that component in standby mode.
• When the first component fails, the second component is put into operation.
• Find the pdf of the lifetime of the standby system if the components have independent exponentially distributed lifetimes with the same mean .