Project: Space Shuttle

There were two statements made during the analysis of the space shuttle disaster that can be proved using statistics.

Statement #1:  "When the space shuttle was launched, the crew had the same chances of surviving as if they were playing Russian Roulette using a gun having 8 chambers and 1 bullet."

Compare Russian Roulette to the launch of the space shuttle.
Russian Roulette involves putting one bullet into a gun having 8 chambers, spinning the chambers, putting the gun to your head and pulling the trigger.
A.  Calculate for Russian Roulette:
     1.  P(Surviving Russian Roulette)

           P(Surviving Russian Roulette) = P(Getting an Empty Chamber) = 7 / 8 = 0.875.

     2.  P(Fatal accident playing Russian Roulette)

          P(Fatal accident playing Russian Roulette) = P(Getting a Loaded Chamber) = 
         1 / 8 = 0.125.

B.  Launching of the Space Shuttle
     The O-ring (gasket) keeping the fuel gases from escaping is held in place by 6 field joints. 
     The P(A single field joint functioning) = 0.977.  The 6 field joints act independently of
     each other and all six field joints must function properly to prevent an explosion on lift-off.
     Calculate:
     1.  P(O-ring functioning) which means a safe lift-off.

         P(All six field joints functioning independently) = 0.977 ^ 6 0.870.

     2.  P(O-ring malfunctioning) which means an explosion on lift-off.

          P(O-ring malfunctioning) = 1 - P(O-ring functioning) = 1 - 0.870 0.13 .

C.  Compare the probabilities calculated in problems A and B.

      0.125 is approximately 0.13. 

D.  Is statement #1 true or false?

     Statement #1 is true.

Statement #2:  "When the space shuttle was redesigned, it had an additional O-ring installed that was truly independent of the other two O-rings.  This new O-ring increased the probability of the O-rings functioning successfully."

A.  Using the same O-ring probabilities from the previous problem, fill in the probability
     distribution table for the O-rings.  The initial set of two rings are dependent and act
     as one O-ring (when one O-ring moved, so did the other O-ring.)  The newly installed
     third O-ring is truly independent of the other O-rings.  Treat the third O-ring as a
     second O-ring.

Event	Probability
First set of O-rings function and second O-ring functions	0.87 * 0.87 0.7569
First set of O-rings malfunction and second O-ring functions	0.13 * 0.87 0.1131
First set of O-rings function and second O-ring malfunctions	0.87 * 0.13 0.1131
First set of O-rings malfunction and second O-ring malfunctions	0.13 * 0.13 0.0169

B.  Let x represent the number of O-ring malfunctions.  Use the probabilities from the above
     table to fill in the probability distribution table. 

C.  Use the probability distribution table above to find:
     1.  P(Explosion on lift-off)

         P(Explosion on lift-off) = P(First O-rings malfunction and second O-ring
         malfunctions) = P(2) = 0.0169.              

     2.  P(Safe lift-off)

         P(Safe lift-off) = 1 - P(Explosion on lift-off) = 1 - 0.0169 = 0.9831

OR

          P(Safe lift-off) = P(First set of O-rings function and second O-ring functions) OR
          P(First set of O-rings malfunction and second O-ring functions) OR
          P(First set of O-rings function and second O-ring malfunctions) =
          P(0) + P(1) = 0.7569 + 0.2262 = 0.9831.

D.  Compare the probabilities for a Safe Lift-Off from Statements 1 and 2.

      Statement #1 has P(Safe-lift off, no explosion) 0.870. 
      Statement #2 has P(Safe-lift off, no explosion) 0.9831.

E.  Is statement #2 true or false?

      Statement #2 is true.

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