Question

. 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. 

Answer 1

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 𝑞 = 0.1, 𝑝 = 1 − 𝑞 = 1 − 0.1 = 0.9.

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

The probability mass function (𝑝. 𝑚. 𝑓) is

𝑃(𝑋 = 𝑥) = 𝑏(𝑥; 𝑛, 0.9), where 𝑥 = 0, 1, 2, … , 𝑛

𝑃(𝑋 = 𝑥) = ($n \atop x$) (0.9)^𝑥 (0.1)^𝑛−𝑥 , where 𝑥 = 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~𝐵𝑖𝑛(2, 𝑝 = 0.9).

Then 𝑃(at least one − half of its engines run ) = 𝑃(𝑋 ≥ 1) = 1 − 𝑃(𝑋 = 0) = = 1 − ($2 \atop 0$) (0.9)⁰(0.1)²⁻⁰ = 1 − (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~𝐵𝑖𝑛(4, 𝑝 = 0.9).

Then 𝑃(at least one − half of its engines run ) = 𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 < 2) = 1 − (𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)) = 1 − (($4 \atop 0$) (0.9)⁰(0.1)⁴⁻⁰ + ($4 \atop1$) (0.9)¹(0.1)⁴⁻¹) = 1 − ((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.