K Engine Start Probability Calculator

[ankitj]

I think this is correct. I would appreciate any input.

Question:
The potential buyer of a particular engine requires that the engine start successfully 10 consecutive times. Suppose that the probability of a successful start is .990. Consider that the outcomes of attempted starts is independent. What is the probability that the engine is accepted after just 10 starts?

My solution:

P(x<=10) = p(2) + p(3) +...+ p(10)

where:
p(2) = P(X = 2) = P(S on #1 and S on #2) = p2
p(3) = P(S on #3 and S on #2 and F on #1) = (1 – p)p2
p(4) = P(S on #4 and S on #3 and F on #2) = (1 – p)p2
p(5) = P(S on #5 and S on #4 and F on #3 and no 2 consecutive S’s on trials prior to #3) = [ 1 – p(2) ](1 – p)p2
p(6) = P(S on #6 and S on #5 and F on #4 and no 2 consecutive S’s on trials prior to #4) = [ 1 – p(2) – p(3)](1 – p)p2

In general, for x = 5, 6, 7, …: p(x) = [ 1 – p(2) - … – p(x - 3)](1 – p)p2

Does this make any sense?
AJ

 

[mathman]

I can't make sense of what you wrote. However the probability of 10 successes in a row is simply .99¹⁰.

 

[ankitj]

But what is the question asking for, in terms of probability? That's what I don't understand.

 

[jgens]

The buyer will accept the engine if and only if it starts successfully ten consecutive times. This means that the engine is "accepted" if it can start ten times in a row. The question is essentially asking what is the probability of this event occurring? That's why the probability is simply 0.99¹⁰ as mathman wrote.

 

[ankitj]

hmm ok. Thank you for your help.