Here given,
engine fail probabilityP(F) = 1−p independently from engine to engine.
engine success probability P(S) = 1-(1-p)
= p
There no question specified . I think a question is like
'For what values of p is a four-engine plane preferable to a two-engine plane?'
For making four engine airplane success at least 2 engine of them should be success.
so, probability = (4C2)p2(1−p)2+(4C3)p3(1−p)1+(4C1)p4(1−p)0=6p2(1−p)2+4p3(1−p)+4p4
required Probability for 2 engine airplane success at least 1 engine should success
= (2C1)p1(1−p)1+(2C2)p2(1−p)0
=2(p)(1-p) + p2
The four-engine plane is safe if:
6p2(1−p)2+4p3(1−p)+4p4≥2(p)(1−p)+p2
6p(1−p)+4p(1−p)+p≥2−p
3p3−8p2+7p−2≥0or(p−1)2(3p−2)≥0
3p−2≥0so,p≥2/3
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