I worked this out on the back of a beer coaster........
Assume that airplane engines operate independently of each other and that at least half of the engines on a plane must operate for the plane to continue flying.
A particular airplane engine fails with a probability of 1/7. which is safer, an airplane with 2 of these engines or a airplane with 4 of these engines
Consider the possible scenarios for the engines with a two engine plane.
The engine can either work (P=6/7) or fail (P=1/7).
Let's look at the probability of the independent events.
1.P(W1,W2)=P(W1)*P(W2)=6/7*6/7=36/49
2.P(W1,F2)=P(W1)*P(F2)=6/7*1/7=6/49
3.P(F1,W2)=P(F1)*P(W2)=1/7*6/7=6/49
4.P(F1,F2)=P(F1)*P(F2)=1/7*1/7=1/49
Disaster for a two engine plane is when both engines fail (1/2 or 1 engine would be OK).
That is case 4 and the probability is 1/49.
..
It works the same for a four engine plane but more outcomes (16 instead of 4).
The probabilities are the same : W(6/7), F(1/7), the denominator now is 7*7*7*7=2401.
..
1.W1*W2*W3*W4=6*6*6*6=1296/2401
2.W1*W2*W3*F4=6*6*6*1=216/2401
3.W1*W2*F3*W4=6*6*1*6=216/2401
4.W1*W2*F3*F4=6*6*1*1=36/2401
5.W1*F2*W3*W4=6*1*6*6=216/2401
6.W1*F2*W3*F4=6*1*6*1=36/2401
7.W1*F2*F3*W4=6*1*1*6=36/2401
8.W1*F2*F3*F4=6*1*1*1=6/2401
9.F1*W2*W3*W4=1*6*6*6=216/2401
10.F1*W2*W3*F4=1*6*6*1=36/2401
11.F1*W2*F3*W4=1*6*1*6=36/2401
12.F1*W2*F3*F4=1*6*1*1=6/2401
13.F1*F2*W3*W4=1*1*6*6=36/2401
14.F1*F2*W3*F4=1*1*6*1=6/2401
15.F1*F2*F3*W4=1*1*1*6=6/2401
16.F1*F2*F3*F4=1*1*1*1=1/2401
.
Disaster for a 4 engine plane is when 3 or more engines fail (2 engines failing is OK).
Look for those cases (3 F or more), they are 8,12,14,15,16.
Their probabilities add,
P(3 or more engines failing)=P(8)+P(12)+P(14)+P(15)+P(16)
P(3 or more)=(6+6+6+6+1)/2401=25/2401
.
For the two engine plane, P(disaster)=1/49=0.0204
For the four engine plane, P(disaster)=25/2401=0.0104
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Therefore the four engine plane would be a safer bet but we all know that don't we?
By the way the calculation is even better if the plane has six, eight or ten engines...that is where a cost/benefit calculation is then made,
When does the cost of the airfare outweigh the risk that the passengers are willing to take to get to where they are going?