Using the binomial distribution, it is found that there is a 0.0001 = 0.01% probability that the airplane will be in serious trouble the next time it goes up.
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
The values of the parameters are given as follows:
p = 0.1, n = 2.
The plane is in serious trouble if both engines fail, that is, the probability is P(X = 2), hence:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 2) = C_{2,2}.(0.1)^{2}.(0.9)^{0} = 0.0001[/tex]
0.0001 = 0.01% probability that the airplane will be in serious trouble the next time it goes up.
More can be learned about the binomial distribution at https://brainly.com/question/24863377
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