Recent content by danmanchester

Hi Aleph, thanks for the clarification!

Thanks for the replies guys. So, for one journey the probability that no engines fail is given by: Pr(X = 0) = 4C0X0.9^4 = 0.6561 And so for four journeys, 0.6561^4 = 0.185? This seems a little bit of an unrealistic figure?

Hi all, I've got a past paper question that goes something like this: An aeroplane has four engines. During a certain journey, each engine fails with a probability of 0.1, independantly of the others. The aeroplane can fly when at least two engines are working. Calculate: A) The...