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IMDB rating: 6.00 Plot: Zane Ziminski is fired from NASA after picking up unidentified signals from a distant location in space. His curiosity soon turn to sheer terror as he travels to Mexico, where a series of unexplained events lead him to believe that there are aliens all over the world. He gathers evidence to support this theory but will he ever have the chance to broadcast the ‘facts’ alive? |
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Suppose calls are arriving at a telephone exchange at n average rate of one per second, according to a poisson?
Suppose calls are arriving at a telephone exchange at n 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 calls;
b) the probability that the fourth call arrives by time t=5 seconds,
c) the expected time at which the fourth call arrives.
Please help! Thank you!
a) since the arrivals are poisson distributed, thus the inter arrival times ( which is a continuous random variable) are exponentially distributed with mean 1/n, where n=rate = 1/sec
thus the probability distribution function is F(t)=n*exp(-n*t).
this is a memoryless distribution, thus if you ask with reference to any kth packet and the (k+1)th packet, the answer does not change with k.
the answer is = integration of F(t).dt with limits 0 to 2 = 1 – exp(-2*n) = 1 – exp(-2)
b) poisson distribution is P(q) = exp(-t)*(t)^q/factorial(q) , where t= time interval in question.
thus for t=5 , P(q) = exp(-5)*(5)^q/factorial(q)
( P(q) is the probability of getting q packets in time interval of 5 )
probaility of getting the 4th call within 5 seconds is =
probability of getting 4 or more calls in time of 5
that is P(4) + P(5) + …… = 1 – ( P(0) + P(1) + P(2) + P(3) ) ; put the values and get the answer.
c) thats simply (number of calls by)/rate = 4/n = 4 seconds
anwoy | Nov 20, 2009