Question

Λ=2.87986530093818

P(Z3 = 1|Z1 = 3) = .448
P(Z1 = 1|Z3 = 1) = .097 .Let T be the minimal value of n such that Xn = 0.
It is called the extinction time. If the process survives, we set T.

Answer 1

The desired probability of the extinction time T > 1 is approximately 0.0498. The duration of time between no-hitters, denoted as T, follows an exponential distribution. The number of events X per unit time follows a Poisson distribution with mean λ if the waiting time T between events follows an exponential distribution.

Step-by-step explanation:

The desired probability is P(T > 1) = 1 − P(T < 1) = 1 − (1 – e¯³) = e¯³ ≈ 0.0498.

Let T = duration of time between no-hitters. We find P(T > 2|T > 1), and by the memoryless property this is simply P(T> 1), which we found to be 0.0498 in part a.

If T represents the waiting time between events, and if T ~ Exp(^), then the number of events X per unit time follows the ak e Poisson distribution with mean λ. The probability density function of X is P(X = k) = -λ^k * e^(-λ) / k!