document.write( "Question 163460: 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 \n" ); document.write( "

Algebra.Com's Answer #120432 by Fombitz(32388) $\"\"$ $\"About$

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Let's look at the possible scenarios for the engines with a two engine plane.
\n" ); document.write( "The engine can either work (P=6/7) or fail (P=1/7).
\n" ); document.write( "Let's look at the probability of the independent events.
\n" ); document.write( "1.P(W1,W2)=P(W1)*P(W2)=6/7*6/7=36/49
\n" ); document.write( "2.P(W1,F2)=P(W1)*P(F2)=6/7*1/7=6/49
\n" ); document.write( "3.P(F1,W2)=P(F1)*P(W2)=1/7*6/7=6/49
\n" ); document.write( "4.P(F1,F2)=P(F1)*P(F2)=1/7*1/7=1/49
\n" ); document.write( "Disaster for a two engine plane is when both engines fail (1/2 or 1 engine would be OK).
\n" ); document.write( "That is case 4 and the probability is 1/49.
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\n" ); document.write( "It works the same for a four engine plane but more outcomes (16 instead of 4).
\n" ); document.write( "The probabilities are the same : W(6/7), F(1/7), the denominator now is 7*7*7*7=2401.
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\n" ); document.write( "1.W1*W2*W3*W4=6*6*6*6=1296/2401
\n" ); document.write( "2.W1*W2*W3*F4=6*6*6*1=216/2401
\n" ); document.write( "3.W1*W2*F3*W4=6*6*1*6=216/2401
\n" ); document.write( "4.W1*W2*F3*F4=6*6*1*1=36/2401
\n" ); document.write( "5.W1*F2*W3*W4=6*1*6*6=216/2401
\n" ); document.write( "6.W1*F2*W3*F4=6*1*6*1=36/2401
\n" ); document.write( "7.W1*F2*F3*W4=6*1*1*6=36/2401
\n" ); document.write( "8.W1*F2*F3*F4=6*1*1*1=6/2401
\n" ); document.write( "9.F1*W2*W3*W4=1*6*6*6=216/2401
\n" ); document.write( "10.F1*W2*W3*F4=1*6*6*1=36/2401
\n" ); document.write( "11.F1*W2*F3*W4=1*6*1*6=36/2401
\n" ); document.write( "12.F1*W2*F3*F4=1*6*1*1=6/2401
\n" ); document.write( "13.F1*F2*W3*W4=1*1*6*6=36/2401
\n" ); document.write( "14.F1*F2*W3*F4=1*1*6*1=6/2401
\n" ); document.write( "15.F1*F2*F3*W4=1*1*1*6=6/2401
\n" ); document.write( "16.F1*F2*F3*F4=1*1*1*1=1/2401
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\n" ); document.write( "Disaster for a 4 engine plane is when 3 or more engines fail (2 engines failing is OK).
\n" ); document.write( "Look for those cases (3 F or more), they are 8,12,14,15,16.
\n" ); document.write( "Their probabilities add,
\n" ); document.write( "P(3 or more engines failing)=P(8)+P(12)+P(14)+P(15)+P(16)
\n" ); document.write( "P(3 or more)=(6+6+6+6+1)/2401=25/2401
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\n" ); document.write( "For the two engine plane, P(disaster)=1/49=0.0204
\n" ); document.write( "For the four engine plane, P(disaster)=25/2401=0.0104
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\n" ); document.write( "The four engine plane would be a safer bet.
\n" ); document.write( "You probably want to get some more reliable engines too. \n" ); document.write( "

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