Shocks to a system occur according to a Poisson process withrate = 0.2. The syst

Question

Shocks to a system occur according to a Poisson process withrate = 0.2. The system will fail if it receives twoconsecutive shocks that are less than one time unit apart. Suppose at time 2 the system is observed to be stillfunctioning. (a) If it is known that exactly one shock occurred inthe interval [0,2), what is the probability that the next shockwill cause the system to fail. (b) If it is knwon that exactly two shocks occurred inthe interval [0,2), what is the probability that the next shockwill cause the system to fail. HINT: You need to find the distribution of the time ofthe most recent shock before time 2 given that there were 2 shocksin [0,2) and they were at least 1 time unit apart. Shocks to a system occur according to a Poisson process withrate = 0.2. The system will fail if it receives twoconsecutive shocks that are less than one time unit apart. Suppose at time 2 the system is observed to be stillfunctioning. (a) If it is known that exactly one shock occurred inthe interval [0,2), what is the probability that the next shockwill cause the system to fail. (b) If it is knwon that exactly two shocks occurred inthe interval [0,2), what is the probability that the next shockwill cause the system to fail. HINT: You need to find the distribution of the time ofthe most recent shock before time 2 given that there were 2 shocksin [0,2) and they were at least 1 time unit apart.

Explanation / Answer

= 0.2.
Failure if 2 in one time unit.
one in the interval [0,2),
probability 2 in one time unit? twoin the interval [0,2),
probability 2 in one time unit?

= 1;. x = 2; = 0.2.
Failure if 2 in one time unit.
one in the interval [0,2),
probability 2 in one time unit? twoin the interval [0,2),
probability 2 in one time unit?

= 1;. x = 2; P(x; ) = (e-)(x) / x!
P(2; 1) = (2.71828-1) (12) / 2!
P(2:1) = .1839
Thus, the probability of selling 2 in one unit is 18.4%


= 2;. x = 2; P(x; ) = (e-)(x) / x!
P(2; 2) = (2.71828-2) (22) / 2!
P(2:2) = .27
Thus, the probability of selling 2 in one unit is 27%
= 2;. x = 2; P(x; ) = (e-)(x) / x!
P(2; 2) = (2.71828-2) (22) / 2!
P(2:2) = .27
Thus, the probability of selling 2 in one unit is 27%

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