Question

The number of particles emitted from a radioactive source during a specified period is a random variable with a Poisson distribution. If the probability of no emissions is 1/3, what is the probability that 2 or more emissions occur?

ans: 2−ln 3 / 3 .

Answer 1

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Answer:

Given :X + P(m) Therefoere, the pmf of X is given by, e-mm" P(X = 1) = r! ;x=0,1,........ = 0; otherwise Since P(X = 0) = 1/3 = 0.3333 Therefore, e-mmº P(X = 0) = m O! = 0.3333 e-m *1 1 =0.3333 e-m = 0.3333 Thaking In on both sides ..-m = -1.0986 im = 1.1 Now, we want to find the P(X > 2) Therefore, P(X > 2) = 1- P(X<2) = 1 – [P(X = 1) + P(X = 0)] =1-1-1-1*1.11 - 6-1.1*1.1°, 11 01 = 1 – [0.3662 +0.3333] = 0.3009

Note: You have provided the answer as (2 - Ln(3))/3 which when simplified gives the same answer as 0.300