A company that buys excavators from an subcontractor knows by experience that certain proportion of the excavators will be faulty. 3 types of errors occurs,denoting by errortype a,b and c. Probability that errortype a occurs on a random selected excavator is 6%. For b it is 2% and for c 5%. The errors occurs independently of each other. a) Calculate the probability that a randomly chosen excavator has atleast 1 of the 3 errors b) Assume that one excavatoris shown to have both error a and b. What is probability that it also have error c ? Ans: a)0.875 b)5.0
This assumes that the errors are independent. a) The prob of none of the errors is 0.94X0.98X0.95=0.87514 The probability of at least one error is 1-0.87514=0.12486 b) A probability of 5.0 is not possible The answer is 0.05 since the errors are assumed to be independent.
P(a) = 0.06 => P(not a) = 1 - P(a) = 0.94 P(b I not a) = 0.94*0.02 = 0.0188 => P(not(b I not a)) = 1 - P(b I not a) = 0.9812 P(c I not(b I not a)) = 0.9812*0.05 = 0.04906 Now the probability that a randomly chosen excavator has atleast 1 of the 3 errors P(a) + P(b I not a) + P(c I not(b I not a)) = 0.06 + 0.0188 + 0.04906 = 0.12786 And the probability that a randomly chosen excavator has no error is 1 - 0.12786 = 0.87214 b) this is obvious: 0.05 since c error is independent.
a) 1 - (0.06 + 0.02 + 0.05) = 1 - 0.13 = 0.87............. NOT 0.875 b) 0.05................................. NOT 5.0 --------------------------------------...
