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

The probability that there will be

(a) at least 2 such accidents in the next month?

= 0.864

(b) at most 1 accident in the next month?

= 0.136

Step-by-step explanation:

This is a Poisson Distribution question.

The formula is given as:

For X = x ,

= λ^x × e^-λ/x!

The monthly worldwide average number of air- plane crashes of commercial airlines is 3.5. What is the probability that there will be

(a) at least 2 such accidents in the next month?

= λ^x × e^-λ/x!

λ = 3.5

x = at least 2 such accidents

x ≥ 2

= 1 - [P(x = 0) + P(x = 1)]

= 1 - [(3.5^0 × e^-3.5/0!)+ (3.5^1 × e^-3.5/1!)]

= 1 - [ 1 × 0.0301973834 + 3.5 × 0.0301973834]

1 - [ 0.0301973834+ 0.105690842]

= 0.8641117746

≈ 0.864

(b) at most 1 accident in the next month?

= λ^x × e^-λ/x!

λ = 3.5

x = at most 1 such accidents

x ≤ 1

= [P(x = 0) + P(x = 1)]

= [(3.5^0 × e^-3.5/0!)+ (3.5^1 × e^-3.5/1!)]

= [ (1 × 0.0301973834) + (3.5 × 0.0301973834)]

[ 0.0301973834+ 0.105690842]

= 0.1358882254

≈ 0.136

the answer is c)8

step-by-step explanaction: