3 ms·
> Based on current engine failure rate, that's (17/18)^9 + (1/18)^8 + C(9,2) * (1/18)^2 * (17/18)^7 I think the second part of that calculation is incorrect. I
by ordinary 14y ago
> Based on current engine failure rate, that's (17/18)^9 + (1/18)^8 + C(9,2) * (1/18)^2 * (17/18)^7
I think the second part of that calculation is incorrect. I'm getting:
P(0 engine fail) = (17/18)^9 ~= 0.5978
P(1 engine fail) = (17/18)^8 * (1/18) * C(9,1) ~= 0.3165
P(2 engine fail) = (17/18)^7 * (1/18)^2 * C(9,2) ~= 0.0745
Adding those together gives:
P(<=2 engine fail) = (17/18)^9 + (17/18)^8 / 2 + (17/18)^7 / 9 ~= 0.9888
That's the same answer you came up with, so I guess you have the correct calculation written down somewhere. Moving to 3 or more engine fails then gives:
P(>=3 engine fail) = 1 - P(<=2 engine fail) ~= 0.0112
- aeontech 14y agoyou're absolutely right, i was doing the final subtraction in my head and not paying attention...