1. ## Probability...again

i'm still having trouble grasping probability. Can anyone help me with this one...An aircraft is equipped with three engines that operate independently. The probability of an engine failure is .01. What is the probability of a successful flight if only one engine is needed for the successful operation of the aircraft?

I doubt this is right but i had 3C2*.99*.01^2 which is .000297 and i subtracted that from one to get .999703. Can anyone help me?

2. Try the case in which all three do not fail.
$1 - (.01)^3$.
That is at least one does not fail.

3. ## ?

i don't get it

4. ^what to do...... do i factor in the probability of all three working, 2 working, a*d 1 working. So 3C3*.99^+3C2*.99^2*.01+3C1*.99*.01^2 which is .999999. Can someone explain?

5. Originally Posted by jarny
^what to do...... do i factor in the probability of all three working, 2 working, a*d 1 working. So 3C3*.99^+3C2*.99^2*.01+3C1*.99*.01^2 which is .999999. Can someone explain?
what is the probability of all three failing?

we have all three angines failing if: the first engine fails and the second engine fails and the third engine fails

since all have to happen at the same time, we multiply the probabilities. that is, the probability of all three engines failing is: $0.01(0.01)(0.01) = (0.01)^3$

this is the ONLY way for the flight to NOT be successful. since the flight must either be successful or not, one or the other must happen and therefore it is a certainty and the probabilities must add up to one.

that is: probability of successful flight + probability of unsuccessful flight = 1 .......since one will certainly happen, the probability of a certain event is 1

that means probability of successful flight = 1 - probability of unsuccessful flight = $1 - (0.01)^3$

This is my 19th post!!!!!

6. oh alright. I kinda was overthinking it then. Thanks a lot both of you! congrats on the 1900th post