# Another question!

1. Aug 12, 2009

### 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

2. Aug 12, 2009

### mathman

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

3. Aug 12, 2009

### ankitj

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

4. Aug 12, 2009

### 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 occuring? That's why the probability is simply 0.9910 as mathman wrote.

5. Aug 13, 2009

### ankitj

hmm ok. Thank you for your help.