I think this is correct. I would appreciate any input.(adsbygoogle = window.adsbygoogle || []).push({});

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

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