Answer :
Answer:
The answer is p ≥ [tex]\frac{1}{2}[/tex]
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
= [tex]\binom{5}{3}p^3(1-p)^2 + \binom{5}{4}p^4(1-p) + \binom{5}{5}p^5[/tex]
= [tex]10p^3 - 20 p^4 + 10p^5 + 5p^4 - 5p^5 + p^5[/tex]
= [tex]6p^5 - 15p^4 + 10p^3[/tex]
We find the probability for a successful flight for a 3 plane engine is given by
[tex]= \binom{3}{2} p^2(1 - p) + \binom{3}{3} p^3[/tex]
= [tex]3p^2 - 3p^3 + p^3[/tex]
= [tex]-2p^3 + 3p^2[/tex]
A 5 engine plane will be preferable to a 3 engine plane
[tex]6p^5 - 15p^4 + 10p^3 \geq -2p^3 + 3p^2[/tex]
[tex]6p^5 - 15p^4 + 12p^3 -3p^2 \geq 0[/tex]
[tex]3p^2 (p-1)^2 (2p -1 ) \geq 0[/tex]
Solving the above inequality we get p ≥ [tex]\frac{1}{2}[/tex] because p cannot be greater than 1 and p greater than 0 does not really signify much.
The answer is p ≥ [tex]\frac{1}{2}[/tex]