Question

Suppose that, in flight, airplane engines will fail with probability 1 − p, independently from engine to engine. If an airplane needs a majority of its engines operative to complete a successful flight, for what values of p is a 5-engine plane preferable to a 3-engine plane?

Answer 1

Answer:

The answer is p ≥  $\frac{1}{2}$

Step-by-step explanation:

The probability of failure of an engine is 1 - p.

The probability that the engine operates successfully  is  = 1 - (1 - p) = p

For a successful flight a majority of its engine have to operate.

So for a 5 engine plane to have a successful flight at least 3 of the engines have to operate successfully.

For a 3 engine plane at least 2 of its engines  have to operate for the total flight for the flight to be successful.

Therefore using the binomial distribution we find the probability for a successful flight for a 5 plane engine is given by

= $\binom{5}{3}p^3(1-p)^2 + \binom{5}{4}p^4(1-p) + \binom{5}{5}p^5$

= $10p^3 - 20 p^4 + 10p^5 + 5p^4 - 5p^5 + p^5$

= $6p^5 - 15p^4 + 10p^3$

We find the probability for a successful flight for a 3 plane engine is given by

$= \binom{3}{2} p^2(1 - p) + \binom{3}{3} p^3$

= $3p^2 - 3p^3 + p^3$

= $-2p^3 + 3p^2$

A 5 engine plane will be preferable to a 3 engine plane

$6p^5 - 15p^4 + 10p^3 \geq -2p^3 + 3p^2$

$6p^5 - 15p^4 + 12p^3 -3p^2 \geq 0$

$3p^2 (p-1)^2 (2p -1 ) \geq 0$

Solving the above inequality we get p ≥  $\frac{1}{2}$  because  p cannot be greater than 1 and p greater than 0 does not really signify much.

The answer is p ≥  $\frac{1}{2}$