trigonometry probability problem, please show works

Question

A Boeing 707 has 4 jet engines. Assume that the probability of any one engine failing during flight is 0.1.(It is not nearly that high, but assume it anyway)

a. What is the probability that a given engine does not fail?

b. Calculate the probabilities that 0,1,2,3, and 4 of the engines fail during a given flight

c. Show that the probabilities in part b add up to 1

d. If the plane will keep flying as long as no more than one engine fails, what is the probability that the plane keeps flying?

e. A Boeing 717 has only 3 jet engines. If the plane will keep flying with no more than 1 engine out, and the probability of any one engine failing is the same as in part a, above, what is the probability that a 727 keeps flying?

f. Based on these calculations, which is safer, a 3-engine plane or a 4-engine plan?

please show works instead of just the answers

Answer 1

a. If the probability of an engine failing is .10, then the compliment of this would be 1-.10 = .90 This

means that there is a probability of .90 that the engine does not fail.

b. This question involves the Binomial Probablility Distribution. In order to calculate the probability of each

event of engines failing we first need to look at the formula and then assign our variables.

The formula for the Binomial Probability is:

P(X = x) = _nC_xp^xq^n-x

n = 4 the total number of engines

x = the number of engines that fail (this is what we call a success)

p = .10 this is the probability of an engine failing

q = .90 this is the probability of an engine not failing

P(X = 0) = ₄C₀(.10)⁰(.90)^4-0 = .6561

P(X = 1) = ₄C₁(.10)¹(.90)^4-1 = .2916

P(X = 2) = ₄C₂(.10)²(.90)^4-2 = .0486

P(X = 3) = ₄C₃(.10)³(.90)^4-3 = .0036

P(X = 4) = ₄C₄(.10)⁴(.90)^4-4 = .0001

c. When we add the probabilities from part b:

.6561 + .2916 + .0486 + .0036 + .0001 = 1.0000

d. P(X < 2) = P(X = 0) + P(X = 1) = .6561 + .2916 = .9477

e. n = 3

P(X < 2) = P(X = 0) + P(X = 1) = .729 + .243 = .972

f. The 3 engine plain is safer because there is a higher probability that no more than one engine will fail.