The number of accidents occurring in a plant in a month follows Poisson distribution with mean as 5.2. The probability of occurrence of less than 2 accidents in the plant during a randomly selected month is A. 0.029 B. 0.034 C. 0.039 D. 0.044

0.029
0.034
0.039
0.044

The correct answer is $\boxed{\text{B) }0.034}$.

The probability of a Poisson distribution with mean $\mu$ of less than $k$ events is given by the following formula:

$$P(X < k) = \sum_{i=0}^{k-1} \frac{\mu^i e^{-\mu}}{i!}$$

In this case, $\mu = 5.2$ and $k = 2$. Substituting these values into the formula, we get:

$$P(X < 2) = \sum_{i=0}^{1} \frac{5.2^i e^{-5.2}}{i!} = 0.034$$

The other options are incorrect because they do not correspond to the correct probability.

Option A, $0.029$, is the probability of a Poisson distribution with mean $\mu = 5.2$ of exactly one event.

Option C, $0.039$, is the probability of a Poisson distribution with mean $\mu = 5.2$ of exactly two events.

Option D, $0.044$, is the probability of a Poisson distribution with mean $\mu = 5.2$ of three or more events.

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