A particular pavement is known to fail from cracking or rutting or both. The pav
ID: 3329850 • Letter: A
Question
A particular pavement is known to fail from cracking or rutting or both. The pavement is forecasted to have medium truckload traffic with a probability of 0.3 and high truckload traffic with a probability of 0.7. If the pavement has medium truckload traffic, then the probability of cracking is 0.02 and rutting is 0.005. If the pavement has high truckload traffic, then the probability of cracking is 0.04 and rutting is 0.008. If the pavement has rutting, then the probability of cracking is 0.6. Determine the probability of pavement failure Hint: Determine in the following order: (i) probability of cracking, (ii) probability of rutting, (iii probability of cracking and rutting. With these probabilities you should be able to determine probability of failure.]Explanation / Answer
Here we are given that:
P( medium ) = 0.3 and P( high ) = 0.7
Also we are given that: P ( crack | medium ) = 0.02 and P( rutting | medium ) = 0.005
P( crack | high ) = 0.04 and P( rutting | high ) = 0.008
Also we are given that P ( crack | rutting ) = 0.6
Using law of total probability, we get:
P( crack ) = P( crack | medium ) P( medium ) + P( crack | high ) P( high )
P( crack ) = 0.02*0.3 + 0.04*0.7 = 0.034
Also,
P( rutting ) = P( rutting | medium ) P( medium ) + P( rutting | high ) P( high )
P( rutting) = 0.005*0.3 + 0.008*0.7 = 0.0071
Now we are given that, P ( crack | rutting ) = 0.6, therefore using bayes theorem, we get:
P( crack and rutting ) = P ( crack | rutting )P( rutting ) = 0.6*0.0071 = 0.00426
Now using the law of addition, we get:
P( crack and rutting ) = P( crack ) + P ( rutting ) - P( crack and rutting )
Putting all the values, we get:
P( crack and rutting ) = 0.034 + 0.0071 - 0.00426 = 0.03684
Therefore the required probability here is 0.03684
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