According to a study conducted by a university, children who are injured three or more times tend to sustain these injuries during a relatively limited time, usually 1 year or less. If the average number of injuries per year for school-age children is three, what are the probabilities of these events? (Round your answers to three decimal places.)

(1) The probability that a school-age child will sustain 2 injuries in a year is 0.221.

(2) The probability that a school-age child will sustain at least 2 injuries in a year is 0.0804.

(3) The probability that a school-age child will sustain at most 2 injuries in a year is 0.417.

Step-by-step explanation:

Let X = number of injuries per year for school-age children.

The average number of injuries per year for school-age children is, λ = 3.

The random variable X follows a Poisson distribution with parameter λ.

The probability mass function of X is:

(1)

Compute the probability that a school-age child will sustain 2 injuries in a year as follows:

Thus, the probability that a school-age child will sustain 2 injuries in a year is 0.221.

(2)

Compute the probability that a school-age child will sustain at least 2 injuries in a year as follows:

P (X ≥ 2) = 1 - P (X < 2)

             = 1 - P (X = 0) - P (X = 1)

Thus, the probability that a school-age child will sustain at least 2 injuries in a year is 0.0804.

(3)

Compute the probability that a school-age child will sustain at most 2 injuries in a year as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

Thus, the probability that a school-age child will sustain at most 2 injuries in a year is 0.417.

the answer for this question is 38

-2

7 - 9 = -2

reversed:

-2 + 9 = 7