Probability with wing designs and Baye's Theorem (with my work shown)

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bry32321

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Past experience shows that about 75% of all full-scale tested wing designs are acceptable, which means that they can withstand the stress of normal ‡flight. Wind tunnel testing of small-scale models reduces development cost. 90% of all acceptable wing designs also pass the (small-scale) wind tunnel test. Unfavorable wind tunnel data were obtained on 95% of all wings, that fail in full-scale testing. Find the probability that a randomly selected wing design
(a) passes both, wind tunnel and full-scale testing.
(b) passes the wind tunnel but fails the full-scale testing.
(c) passes the wind tunnel test.
(d) Calculate the probability that a wing design which passed the wind tunnel test will also pass full scale
testing.

My work:
a.) My thoughts were that the probability of this would be = 90%(the ones that passed wind tunnel & full scale) of 75%(the ones that passed full-scale)= (90/100)(75/100)=27/40=0.675

b.) 5%(the ones that passed wind tunnel but failed full scale) of 25%(the ones that failed full scale)=(5/100)(25/100)=1/80=0.0125

c.)P(passes wind tunnel)=(0.675)+(0.0125)=0.6875

d.) (Using Baye's Theorem), I came up with P(passed full scale given passed wind scale) = [P(passed wind tunnel given passed full scale)*P(passed full scale)]/[P(passed wind tunnel)]=[(90%)(75%)]/[(75%)(90%)+(25%)(5%)]=54/55=0.982



Did I do this right?

Thank you for any feedback given!
 
Last edited:
bry32321 said:
Past experience shows that about 75% of all full-scale tested wing designs are acceptable, which means that they can withstand the stress of normal ‡flight. Wind tunnel testing of small-scale models reduces development cost. 90% of all acceptable wing designs also pass the (small-scale) wind tunnel test. Unfavorable wind tunnel data were obtained on 95% of all wings, that fail in full-scale testing. Find the probability that a randomly selected wing design
(a) passes both, wind tunnel and full-scale testing.
(b) passes the wind tunnel but fails the full-scale testing.
(c) passes the wind tunnel test.
(d) Calculate the probability that a wing design which passed the wind tunnel test will also pass full scale
testing.

My work:
a.) My thoughts were that the probability of this would be = 90%(the ones that passed wind tunnel & full scale) of 75%(the ones that passed full-scale)= (90/100)(75/100)=27/40=0.675

b.) 5%(the ones that passed wind tunnel but failed full scale) of 25%(the ones that failed full scale)=(5/100)(25/100)=1/80=0.0125

c.)P(passes wind tunnel)=(0.675)+(0.0125)=0.6875

d.) (Using Baye's Theorem), I came up with P(passed full scale given passed wind scale) = [P(passed wind tunnel given passed full scale)*P(passed full scale)]/[P(passed wind tunnel)]=[(90%)(75%)]/[(75%)(90%)+(25%)(5%)]=54/55=0.982



Did I do this right?

Thank you for any feedback given!
Click to expand...
To apply Bayes's Theorem, we have to express the given statistics as conditional probabilities. Lets use some abbreviations:
Let F = "passes full-scale test," /F = not-F = "fails full-scale test."
.....S = "passes small-scale windtunnel," /S = not-S = "fails small-scale windtunnel."

The given data are
.....P(F) = 0.75, from which P(/F) = 0.25
.....P(S | F) = 0.90............P(/S | F) = 0.10
.....P(/S | /F) = 0.95..........P(S | /F) = 0.05

a) P(F & S) = P(S | F)*P(F) = 0.675...OK

b) P(S & /F) = P(S | /F)*P(/F) = 0.0125...OK

c) P(S) = P(S | F) + P(S | /F) = 0.6875...OK

d) P(F | S) = P(S | F)*P(F)/P(S) = 0.9818...OK

Did fine. Notice that Mr. Bayes had as "s" in his name.
 
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