Suppose calls are arriving at a telephone exchange at an average rate of one per second, according to a poisson arrival process. find: a) the probability that the fourth call after time t = 0 arrives within 2 seconds of the third call; b) the probability that the fourth call arrives by time t = 5 seconds; c) the expected time at which the fourth call arrives.

Explanation has been given below

Step-by-step explanation:                    

a) inter arrival times are exponentially distributed with mean 1/n , where n = rate = 1/sec.                                                                              

  probability distribution function is F(t)=n*exp(-n*t).  

reference to any kth packet and the (k-1)th packet

the answer is = integration of F(t).dt with limits 0 to 2 = 1 - exp(-2*n) = 1 - exp(-2)  

b)  t=5 , P(q) = exp(-5)*(5)^q/factorial(q)  

probability of fourth call within t=5 seconds is =  

that is P(4)   P(5)    = 1 - ( P(0)   P(1)   P(2)   P(3) ) ;  put the values and get the answer.  

c) number of calls/rate =  4/n = 4 seconds

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form