Respuesta :
Answer:
I'm going to make an uneducated guess here.
I think the probability of two engines failing would be the same for either aircraft. In the instance that does happen, the helicopter doesn't fly at all and the airplane does.
I say the airplane is safer.
Step-by-step explanation:
So, the plane is safer than the helicopter.
Binomial distribution:
The formula for finding the probability by Binomial distribution is,
[tex]P(X = x) = n_C_x \times p^x \times (1 - p)^{(n - x)}[/tex]
For Helicopter,
We need to calculate, Probability of safe landing, [tex]P(X = 0)[/tex]
Here, [tex]n = 2, p = 0.99, (1 - p) = 0.01\ and\ x = 1[/tex]
As per binomial distribution formula [tex]P(X = x) = n_C_x \times p^x \times (1 - p)^{(n - x)}[/tex]
We need to calculate [tex]P(X \ge 1).[/tex]
[tex]P(X \ge1) = (2_C_1 \times 0.99^1 \times 0.01^1) + (2_C_2 \times 0.99^2 \times 0.01^0)\\P(X \ge1) = 0.0198 + 0.9801\\P(X \ge 1) = 0.9999[/tex]
For Plane,[tex]P(safe\ landing)[/tex]
Here, [tex]n = 4, p = 0.99, (1 - p) = 0.01\ and\ x = 2[/tex]
As per binomial distribution formula [tex]P(X = x) = n_C_x \times p^x \times (1 - p)^{(n - x)}[/tex]
We need to calculate P(X >= 2).
[tex]P(X \ge 2) = (4_C_2 \times 0.99^2 \times 0.01^2) + (4_C_3 \times 0.99^3 \times 0.01^1) + (4_C_4 \times 0.99^4 \times 0.01^0)\\P(X \ge 2) = 0.0006 + 0.0388 + 0.9606\\P(X \ge2) = 1\\[/tex]
Learn more about the topic Binomial distribution:
https://brainly.com/question/26096510
