Question #72635

Suppose that airplane engines operate independently
and fail with probability equal to 0.4. Assuming
that a plane makes a safe flight if at least one-half of its
engines run, determine whether a 4-engine plane or a 2-
engine plane has the higher probability for a successful
flight.
1

Expert's answer

2018-01-26T08:17:07-0500

Answer on Question #72635 – Math – Statistics and Probability Question

Suppose that airplane engines operate independently and fail with probability equal to 0.4. Assuming that a plane makes a safe flight if at least one-half of its engines run, determine whether a 4-engine plane or a 2-engine plane has the higher probability for a successful flight.

Solution

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

Then q=0.4,p=1q=10.4=0.6q = 0.4, p = 1 - q = 1 - 0.4 = 0.6.

Trials are independent.

Hence, XBin(n,p=0.6)X \sim \text{Bin}(n, p = 0.6)

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


P(X=x)=b(x;n,0.6), where x=0,1,2,,nP(X = x) = b(x; n, 0.6), \text{ where } x = 0, 1, 2, \dots, nP(X=x)=(nx)(0.6)x(0.4)nx, where x=0,1,2,,nP(X = x) = \binom{n}{x} (0.6)^x (0.4)^{n-x}, \text{ where } x = 0, 1, 2, \dots, n


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

If n=2,XBin(2,p=0.6)n = 2, X \sim \text{Bin}(2, p = 0.6).

Then


P(at least onehalf of its engines run)=P(X1)=1P(X=0)==1(20)(0.6)0(0.4)20=1(0.4)2=0.84\begin{array}{l} P(\text{at least one} - \text{half of its engines run}) = P(X \geq 1) = 1 - P(X = 0) = \\ = 1 - \binom{2}{0} (0.6)^0 (0.4)^{2-0} = 1 - (0.4)^2 = 0.84 \end{array}


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,XBin(4,p=0.6)n = 4, X \sim \text{Bin}(4, p = 0.6).

Then


P(at least onehalf of its engines run)=P(X2)=1P(X<2)==1(P(X=0)+P(X=1))==1((40)(0.6)0(0.4)40+(41)(0.6)1(0.4)41)==1((0.4)4+4(0.6)(0.4)3)=0.8208\begin{array}{l} P(\text{at least one} - \text{half of its engines run}) = P(X \geq 2) = 1 - P(X < 2) = \\ = 1 - \left(P(X = 0) + P(X = 1)\right) = \\ = 1 - \left(\binom{4}{0} (0.6)^0 (0.4)^{4-0} + \binom{4}{1} (0.6)^1 (0.4)^{4-1}\right) = \\ = 1 - ((0.4)^4 + 4(0.6)(0.4)^3) = 0.8208 \end{array}


Since 0.84>0.82080.84 > 0.8208, the 2-engine plane has a higher probability for a successful flight than the 4-engine plane.

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

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