The monthly worldwide average number of airplane crashes of commercial airlines is 3.5. What is the probabil- ity that there will be (a) at least 2 such accidents in the next month; (b) at most 1 accident in the next month

Step-by-step explanation:

We would apply the formula for poisson distribution which is expressed as

P(x = r) = (e^- µ × µ^r)/r!

Where

µ represents the mean of the theoretical distribution.

r represents the number of successes of the event.

From the information given,

µ = 3.5

a)

For the probability that there will be at least 2 such accidents in the next month, it is expressed as

P(x ≥ 2) = 1 - P(x < 2)

P(x < 2) = P(x = 0) + P(x = 1)

Therefore,

P(x = 0) = (e^- 3.5 × 3.5^0)/0!

P(x = 0) = (e^- 3.5 × 1)/1

P(x = 0) = 0.03

P(x = 1) = (e^- 3.5 × 3.5^1)/1!

P(x = 1) = (e^- 3.5 × 3.5)/1

P(x = 1) = 0.11

P(x < 2) = P(x = 0) + P(x = 1) = 0.03 + 0.11 = 0.14

P(x ≥ 2) = 1 - 0.14 = 0.86

b)

For the probability that there will be at most 1 accident in the next month, it is expressed as

P(x ≤ 1) = P(x = 0) + P(x = 1)

P(x = 0) = 0.03

P(x = 1) = 0.11

P(x ≤ 1) = 0.03 + 0.11 = 0.14

donde?

Step-by-step explanation: