Answer to Question #180710 in Statistics and Probability for phemelo mongae

Question #180710

. Airplane engines operate independently in flight and fail with probability of 0.1. A plane makes a successful flight if at most half of its engines fail. Determine the probability of a successful flight for two-enginned and four-engined planes.Β 


1
Expert's answer
2021-04-14T08:07:28-0400

Let 𝑋 be the random variable representing the number of engines running out of 𝑛 engines in a plane. Let us consider, running of an engine is a success.

Then π‘ž = 0.1, 𝑝 = 1 βˆ’ π‘ž = 1 βˆ’ 0.1 = 0.9.

Trials are independent. Hence, 𝑋~𝐡𝑖𝑛(𝑛, 𝑝 = 0.9)

The probability mass function (𝑝. π‘š. 𝑓) is

𝑃(𝑋 = π‘₯) = 𝑏(π‘₯; 𝑛, 0.9), where π‘₯ = 0, 1, 2, … , 𝑛

𝑃(𝑋 = π‘₯) = ("n \\atop x") (0.9)π‘₯ (0.1)π‘›βˆ’π‘₯ , where π‘₯ = 0, 1, 2, … , 𝑛

a) For the 2-engine plane to make a successful flight, at least one engine must be running.

If 𝑛 = 2, 𝑋~𝐡𝑖𝑛(2, 𝑝 = 0.9).

Then 𝑃(at least one βˆ’ half of its engines run ) = 𝑃(𝑋 β‰₯ 1) = 1 βˆ’ 𝑃(𝑋 = 0) = = 1 βˆ’ ("2 \\atop 0") (0.9)0(0.1)2βˆ’0 = 1 βˆ’ (0.1)2 = 0.99

b) On the other hand, for the 4-engine plane to make a successful flight, at least two engines must be running.

If 𝑛 = 4, 𝑋~𝐡𝑖𝑛(4, 𝑝 = 0.9).

Then 𝑃(at least one βˆ’ half of its engines run ) = 𝑃(𝑋 β‰₯ 2) = 1 βˆ’ 𝑃(𝑋 < 2) = 1 βˆ’ (𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)) = 1 βˆ’ (("4 \\atop 0") (0.9)0(0.1)4βˆ’0 + ("4 \\atop1") (0.9)1(0.1)4βˆ’1) = 1 βˆ’ ((0.1)4 + 4(0.9)(0.1)3) = 0.9963

Since 0.9963 > 0.99, the 4-engine plane has a higher probability for a successful flight than the 2-engine plane.

Answer: the 4-engine plane has a higher probability for a successful flight than the 2-engine plane.


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