5 pair of rvs
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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.

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5 1 joint pmf
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
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
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
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
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
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
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 .
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