Answer to Question #91658 in Statistics and Probability for Sibel Gercek

Question

2.119) A producer of a certain type of electronic component ships to suppliers inlots of twenty. Suppose that 60% of all such lots contain no defective components,30% contain one defective component, and 10% contain two defective components.A lot is picked, two components from the lot are randomly selected and tested.

a) If both are non-defective, what is the conditional probability that one defec-tive component exist in the lot?
b) If exactly one of the two components is defective, what is the conditionalprobability that two defective components exist in the lot?

Answer 1

Let

$$H_1 - {contain \;no \;defective\; components}$$

$$H_2 - {contain \;one \;defective\; component}$$

$$H_3 - {contain \;two \;defective\; components}$$

$$P(H_1)=0,6; \;\;\;\;P(H_2)=0,3; \;\;\;\;P(H_3)=0,1$$

a)

$$A - {both\; are\; non-defective}$$

$$P(H_2\backslash A)= \frac{P(H_2)P(A\backslash H_2)}{P(A)}=\frac{P(H_2)P(A\backslash H_2)}{P(H_1)P(A\backslash H_1)+P(H_2)P(A\backslash H_2)+P(H_3)P(A\backslash H_3)}$$

$P(A\backslash H_1)=1;$ $P(A\backslash H_2)=\frac{19}{20}*\frac{18}{19}=\frac{18}{20}=0,9;$ $P(A\backslash H_3)=\frac{18}{20}*\frac{17}{19}=0,8.$

Then

$$P(H_2\backslash A)=\frac{0,3*0,9}{1*0,6+0,3*0,9+0,1*0,8}=\frac{0,27}{0,95}=0,28$$

Answer:$P(H_2\backslash A)=0,28.$

b)

$$B-{exactly\; on\; of\; the \; two\; components\; is\; defective}$$

$$P(H_3\backslash B)= \frac{P(H_3)P(B\backslash H_3)}{P(B)}=\frac{P(H_3)P(B\backslash H_3)}{P(H_1)P(B\backslash H_1)+P(H_2)P(B\backslash H_2)+P(H_3)P(B\backslash H_3)}$$

$P(B\backslash H_1)=0;\;\;\;\;P(B\backslash H_2)=\frac{1}{20}=0,05;\;\;\;\;P(B\backslash H_3)=\frac{2}{20}=0,1.$

$$P(H_3\backslash B)= \frac{0,1*0,1}{0,6*0+0,3*0,05+0,1*0,1}=\frac{0,01}{0,025}=0,4.$$

Answer: $P(H_3\backslash B)=0,4.$