Question 1013826: Pouches of specialty chemicals are tested for presence of impurities a or b, which render the chemical unfit for use. The fitness for use is tested separately. Suppose 10% of the pouches have impurity a, 5% have impurity b,1% have both impurities a & b. Other pouches are free of impurities. The probabilities of being unfit for use are 1%, 2%, 6%, and 0.1% for pouches with impurity a,impurity b, both impurities a & b, and no impurities respectively. What is the probability of a randomly chosen pouch being found unfit for use?
Answer by mathmate(429)   (Show Source):
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Question:
Pouches of specialty chemicals are tested for presence of impurities a or b, which render the chemical unfit for use. The fitness for use is tested separately. Suppose 10% of the pouches have impurity a, 5% have impurity b,1% have both impurities a & b. Other pouches are free of impurities. The probabilities of being unfit for use are 1%, 2%, 6%, and 0.1% for pouches with impurity a,impurity b, both impurities a & b, and no impurities respectively. What is the probability of a randomly chosen pouch being found unfit for use?
Solution:
When we randomly select one pouch, the outcomes are:
A: has impurity a
B: has impurity b
AB: has both impurities a and b
N: has neither impurity.
We are given
P(AB)=0.01
P(A or AB)=0.10
P(B or AB)=0.05
We conclude that
P(A)=P(A or AB)-P(AB)=0.10-0.01=0.09
P(B)=P(B or AB)-P(B)=0.05-0.01=0.04
Hence
P(N)=1-(P(A)+P(B)+P(AB)=1-(0.09+0.04+0.01)=1-0.14=0.86
Thus the probability of picking
P{A,B,AB,N}={0.09,0.04,0.01,0.86}
Note that the sum of the four probabilities is 1.0 since they are complementary.
Let a, b, ab, n be the events that the chosen pouch is unfit for use, then the corresponding probabilities for unfit pouches are (as given):
P(a,b,ab,n)={0.01,0.02,0.06,0.001}
Therefore the probability of ending up with any unfit pouch is the product of the probability of picking the category of pouch and the corresponding unfit probability, or P(A)*P(a)....
P(A)*P(a)=0.09*0.01=9*10^-4
P(B)*P(b)=0.04*0.02=8*10^-4
P(AB)*P(ab)=0.01*0.06=6*10^-4
P(N)*P(n)=0.85*0.001=8.6*10^-4
Total probability of unfit pouch
=ΣP(X)*P(x)
=(9+8+6+8.6)*10^4
=0.00316
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