Using Binomial distribution, the probability that more than 2 fail-safe components will fail is 0.00154
For given question,
The pmf of a Binomial Distribution is given by,
[tex]f(x) =^nC_x(p)^x(q)^{n-x}[/tex]
where n = number of trials
p = probability of success (success event being missile failing)
q = 1 - p i.e. probability of failure (failure event being missile succeeding)
x = number of successes required (number of failed missiles)
In the given case,
n = 5, p = 0.055,
⇒ q = 0.945
We need to find the probability that more than 2 will fail.
So, x = 3
Using Binomial distribution, the required probability is,
[tex]P(x > 2)\\\\= P(x=3) + P(x=4) + P(x=5)\\\\=^5C_3(0.055)^3(0.945)^2 +~^5C_4(0.055)^4(0.945)+~^5C_5(0.055)^5\\\\ =0.0015 + 0.00004 + 0.0000005\\\\ = 0.00154[/tex]
Therefore, using Binomial distribution, the probability that more than 2 fail-safe components will fail is 0.00154
Learn more about the probability here:
https://brainly.com/question/3679442
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