Answer to Question #219256 in Statistics and Probability for Enam

Question #219256

An item produced by a company is susceptible to two types of defects namely, A and B. The probability that an item has defect A is 1/6 . The probability that it has defect B is 1/8 . Calculate the probability that the item has

(i) Both A and B defects

(ii) Only defect A

(iii) Only defect B

(iv) One defect only

(v) At least one defect


1
Expert's answer
2021-07-21T11:26:08-0400

(i) Since the event AA and event BB are independent, then

P(AB)=P(A)P(B)=16(18)=148P(A\cap B)=P(A)P(B)=\dfrac{1}{6}(\dfrac{1}{8})=\dfrac{1}{48}

(ii)

P(only A)=P(AB)=P(A)P(AB)P(only\ A)=P(A\cap B')=P(A)-P(A\cap B)




=16148=748=\dfrac{1}{6}-\dfrac{1}{48}=\dfrac{7}{48}


(iii)

P(only B)=P(BA)=P(B)P(AB)P(only\ B)=P(B\cap A')=P(B)-P(A\cap B)




=18148=548=\dfrac{1}{8}-\dfrac{1}{48}=\dfrac{5}{48}

(iv)


P(only one)=P(AB)+P(BA)P(only\ one)=P(A\cap B')+P(B\cap A')

=748+548=14=\dfrac{7}{48}+\dfrac{5}{48}=\dfrac{1}{4}

(v)


P(at least one)=P(AB)P(at\ least\ one)=P(A\cup B)

=P(A)+P(B)P(AB)=P(A)+P(B)-P(A\cap B)

=18+16148=1348=\dfrac{1}{8}+\dfrac{1}{6}-\dfrac{1}{48}=\dfrac{13}{48}



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