Answer:
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
