A radioactive source of material emits a radioactive particle with probability 1/100

in each second. Let X be the number of particles emitted in one minute.
1. Determine the probability distribution, the expected value and the variance
of X.
2. Calculate the probability that the number of particles emitted in one minute
is more than 2.

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To determine the probability distribution, expected value, and variance of X, we need to understand that X follows a Poisson distribution. In a Poisson distribution, the average number of events (particles emitted in this case) occurring in a fixed interval is λ.

In this case, the average number of particles emitted in one minute is given by λ = (1/100) * 60 = 0.6.

1. Probability Distribution of X:
The probability mass function (PMF) of a Poisson distribution is given by:
P(X = k) = (e^(-λ) * λ^k) / k!

For X, we obtain:
P(X = k) = (e^(-0.6) * 0.6^k) / k!

2. Expected Value of X:
The expected value (mean) of a Poisson distribution is equal to its parameter λ. So, E(X) = λ = 0.6.

3. Variance of X:
The variance of a Poisson distribution is also equal to its parameter λ. So, Var(X) = λ = 0.6.

Now, let's calculate the probability that the number of particles emitted in one minute is more than 2.

P(X > 2) = 1 - P(X ≤ 2)
To calculate P(X ≤ 2), we need to sum up the probabilities of getting 0, 1, and 2 particles:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

Using the PMF formula, we can substitute k = 0, 1, and 2 to calculate each term.